C H A P T E R 3 Exponential and Logarithmic Functions · 266 Chapter 3 Exponential and Logarithmic Functions 12. Asymptote: x −3 − 2−1 1 2 3 5 4 3 2 −1 y y 5 0 f sx d5 s1

C H A P T E R 3

Exponential and Logarithmic Functions

Section 3.1 Exponential Functions and Their Graphs . . . . . . . . . 265

Section 3.2 Logarithmic Functions and Their Graphs . . . . . . . . 273

Section 3.3 Properties of Logarithms . . . . . . . . . . . . . . . . . 281

Section 3.4 Exponential and Logarithmic Equations . . . . . . . . . 289

Section 3.5 Exponential and Logarithmic Models . . . . . . . . . . 303

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329

Practice Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333

265

C H A P T E R 3

Exponential and Logarithmic Functions

Section 3.1 Exponential Functions and Their Graphs

n You should know that a function of the form where is called an exponential

function with base a.

n You should be able to graph exponential functions.

n You should know formulas for compound interest.

(a) For n compoundings per year:

(b) For continuous compoundings: A 5 Pert.

A 5 P11 1r

n2nt

.

a Þ 1,a > 0,f sxd 5 ax,

Vocabulary Check

1. algebraic 2. transcendental 3. natural exponential; natural

4. 5. A 5 PertA 5 P11 1r

n2nt

1. f s5.6d 5 s3.4d5.6 < 946.852 2. f sxd 5 2.3x 5 2.33/2 < 3.488 3. f s2pd 5 52p < 0.006

4. f sxd 5 s23d5x

5 s23d5s0.3d < 0.544 5.

< 1767.767

gsxd 5 5000s2xd 5 5000s221.5d 6.

< 1.274 3 1025

5 200s1.2d12?24

f sxd 5 200s1.2d12x

7.

Increasing

Asymptote:

Intercept:

Matches graph (d).

s0, 1d

y 5 0

f sxd 5 2x 8. rises to the right.

Asymptote:

Intercept:

Matches graph (c).

s0, 2d

y 5 1

f sxd 5 2x 1 1 9.

Decreasing

Asymptote:

Intercept:

Matches graph (a).

s0, 1d

y 5 0

f sxd 5 22x

10. rises to the right.

Asymptote:

Intercept:

Matches graph (b).

s0, 14d

y 5 0

fsxd 5 2x22 11.

Asymptote: y 5 0x

321−1−2−3

5

4

3

2

1

−1

yf sxd 5 s12dx

x 0 1 2

4 2 1 0.250.5f sxd

2122

266 Chapter 3 Exponential and Logarithmic Functions

12.

Asymptote:

x321−1−2−3

5

4

3

2

−1

y

y 5 0

f sxd 5 s12d2x

5 2x

x 0 1 2

1 2 40.50.25f sxd

2122

13.

Asymptote:

x321−1−2−3

5

4

3

1

−1

y

y 5 0

f sxd 5 62x

x 22 21 0 1 2

36 6 1 0.167 0.028f sxd

14.

Asymptote:

x

5

4

3

321−1−2−3

2

1

−1

y

y 5 0

f sxd 5 6x

x 22 21 0 1 2

0.028 0.167 1 6 36f sxd

15.

Asymptote:

x321−1−2−3

5

4

3

2

1

−1

y

y 5 0

f sxd 5 2x21

x 0 1 2

1 20.50.250.125f sxd

2122

16.

Asymptote: y 5 3

x

7

6

5

4

2

1

1−1−2−3 2 3 4 5

yf sxd 5 4x23 1 3

x 0 1 2 3

43.253.0633.0163.004f sxd

21

17.

Because the graph of g can be obtained

by shifting the graph of f four units to the right.

gsxd 5 f sx 2 4d,

gsxd 5 3x24 f sxd 5 3x, 18.


by shifting the graph of f one unit upward.

gsxd 5 f sxd 1 1,

f sxd 5 4x, gsxd 5 4x 1 1

19.


by shifting the graph of f five units upward.

gsxd 5 5 1 f sxd,

gsxd 5 5 2 2x f sxd 5 22x, 20.

Because the graph of g can be

obtained by reflecting the graph of f in the y-axis and

shifting f three units to the right. (Note: This is equivalent

to shifting f three units to the left and then reflecting the

graph in the y-axis.)

gsxd 5 f s2x 1 3d,

f sxd 5 10x, gsxd 5 102x13

Section 3.1 Exponential Functions and Their Graphs 267

21.


obtained by reflecting the graph of f in the x-axis and

y-axis and shifting f six units to the right. (Note: This is

equivalent to shifting f six units to the left and then

reflecting the graph in the x-axis and y-axis.)

gsxd 5 2f s2x 1 6d,

f sxd 5 s72dx

, gsxd 5 2s72d2x16

22.

hence the graph of g can be obtained

by reflecting the graph of f in the x-axis and shifting the

resulting graph five units upward.

gsxd 5 2f sxd 1 5,

gsxd 5 20.3x 1 5 f sxd 5 0.3x,

27. f s34d 5 e23y4 < 0.472 28. f sxd 5 ex 5 e3.2 < 24.533 29. f s10d 5 2e25s10d < 3.857 3 10222

23.

−3

−1

3

3

y 5 22x224.

−3 3

−1

3

y 5 32|x| 25.

−1

0

5

4

fsxd 5 3x22 1 1 26.

−6 3

−3

3

y 5 4x11 2 2

30.

5 1.5e120 < 1.956 3 1052

f sxd 5 1.5es1y2dx 31. f s6d 5 5000e0.06s6d < 7166.647 32.

5 250e0.05s20d < 679.570

f sxd 5 250e0.05x

33.

Asymptote:

x321−1−2−3

5

4

3

2

1

−1

y

y 5 0

f sxd 5 ex

x 0 1 2

1 7.3892.7180.3680.135f sxd

2122

34.

Asymptote:

x

5

4

3

2

1

−1

321−1−2−3

y

y 5 0

f sxd 5 e2x

x 22 21 0 1 2

7.389 2.718 1 0.368 0.135f sxd

35.

Asymptote:

x−1−2−3−4−5−6−7−8 1

8

7

6

5

4

3

2

1

y

y 5 0

f sxd 5 3ex14

x

31.1040.4060.1490.055f sxd

2425262728

36.

Asymptote:

x

6

5

4

3

2

1

−14321−1−2−3

y

y 5 0

f sxd 5 2e20.5x

x 22 21 0 1 2

5.437 3.297 2 1.213 0.736f sxd


37.

Asymptote:

x7654321

9

8

7

6

5

3

2

1

−1−2−3

y

y 5 4

f sxd 5 2ex22 1 4

x 0 1 2

64.7364.2714.1004.037f sxd

2122

38.

Asymptote:

x

8

7

6

5

4

3

1

87654321−1

y

y 5 2

f sxd 5 2 1 ex25

x 0 2 4 5 6

2.007 2.050 2.368 3 4.718f sxd

45.

x 5 2

x 1 1 5 3

3x11 5 33

3x11 5 27 46.

x 5 7

x 2 3 5 4

2x23 5 24

2x23 5 16 47.

x 5 23

x 2 2 5 25

2x22 5 225

2x22 51

32

48.

x 5 24

x 1 1 5 23

s15dx11

5 s15d23

s15dx11

5 53

s15dx11

5 125 49.

x 513

3x 5 1

3x 1 2 5 3

e3x12 5 e3 50.

x 552

2x 5 5

2x 2 1 5 4

e2x21 5 e4

39.

−7

−1

5

7

y 5 1.0825x 40.

−4 8

−2

6

y 5 1.085x 41.

−10

0

23

22

sstd 5 2e0.12t

42.

−16 17

−2

20

sstd 5 3e20.2t 43.

−3

0

3

4

gsxd 5 1 1 e2x 44.

−2 4

0

4

hsxd 5 ex22

51.

x 5 3 or x 5 21

sx 2 3dsx 1 1d 5 0

x2 2 2x 2 3 5 0

x2 2 3 5 2x

ex223 5 e2x 52.

x 5 3 or x 5 2

sx 2 3dsx 2 2d 5 0

x2 2 5x 1 6 5 0

x2 1 6 5 5x

ex216 5 e5x


58. A 5 Pert 5 12,000e0.06t

53.

Compounded times per year:

Compounded continuously: A 5 Pert 5 2500e0.025s10d

A 5 P11 1r

n2nt

5 250011 10.025

n 210n

n

P 5 $2500, r 5 2.5%, t 5 10 years

n 1 2 4 12 365ContinuousCompounding

A $3200.21 $3205.09 $3207.57 $3209.23 $3210.04 $3210.06

54.

Compounded n times per year:

Compounded continuously: A 5 1000e0.04s10d

A 5 100011 10.04

n 210n

P 5 $1000, r 5 4%, t 5 10 years


A $1480.24 $1485.95 $1488.86 $1490.83 $1491.79 $1491.82

55.

Compounded times per year:

Compounded continuously: A 5 Pert 5 2500e0.03s20d

A 5 P11 1r

n2nt

5 250011 10.03

n 220n

n

P 5 $2500, r 5 3%, t 5 20 years


A $4515.28 $4535.05 $4545.11 $4551.89 $4555.18 $4555.30

56.

Compounded n times per year:

Compounded continuously: A 5 1000e0.06s40d

A 5 100011 10.06

n 240n

P 5 $1000, r 5 6%, t 5 40 years


A $10,285.72 $10,640.89 $10,828.46 $10,957.45 $11,021.00 $11,023.18

57. A 5 Pert 5 12,000e0.04t

t 10 20 30 40 50

A $17,901.90 $26,706.49 $39,841.40 $59,436.39 $88,668.67

t 10 20 30 40 50

A $21,865.43 $39,841.40 $72,595.77 $132,278.12 $241,026.44


59. A 5 Pert 5 12,000e0.065t

t 10 20 30 40 50

A $22,986.49 $44,031.56 $84,344.25 $161,564.86 $309,484.08

60. A 5 Pert 5 12,000e0.035t

t 10 20 30 40 50

A $17,028.81 $24,165.03 $34,291.81 $48,662.40 $69,055.23

64.

(a)

(b) When

(c) Since is on the graph in part (a), it

appears that the greatest price that will still yield a

demand of at least 600 units is about $350.

s600, 350.13d

p 5 500011 24

4 1 e20.002s500d2 < $421.12

x 5 500:

0

0

2000

1200

p 5 500011 24

4 1 e20.002x2 65.

(a)

(b)

(c) Vs2d < 1,000,059.63 computers

Vs1.5d < 10,004.472 computers

Vs1d < 10,000.298 computers

Vstd 5 100e4.6052t

61.

< $222,822.57

A 5 25,000es0.0875ds25d 62.

< $212,605.41

A 5 5000es0.075ds50d 63. Cs10d 5 23.95s1.04d10 < $35.45

66. (a)

Since the growth rate is negative,

the population is decreasing.

(b) In 1998, and the population is given by

In 2000, and the population is given by

(c) In 2010, and the population is given by

Ps20d 5 152.26e20.0039s20d 5 140.84 million.

t 5 20

Ps10d 5 152.26e20.0039s10d 5 146.44 million.

t 5 10

Ps8d 5 152.26e20.0039s8d 5 147.58 million.

t 5 8

20.0039 5 20.39%,

P 5 152.26e20.0039t 67.

(a)

(b)

(c)

0 5000

0

30

Qs1000d < 16.21 grams

Qs0d 5 25 grams

Q 5 25s12dty1599

68.

(a)

(b)

< 7.85 grams

When t 5 2000: Q 5 10s12d2000y5715

5 10s1d 5 10 grams

When t 5 0: Q 5 10s12d0y5715

Q 5 10s12d ty5715

(c)

Time (in years)

t80004000

2

4

6

8

10

12

Q

Mas

s of

14C

(in

gra

ms)


69.

(a)

0

0 120

110

y 5100

1 1 7e20.069x

71. True. The line is a horizontal asymptote for the

graph of f sxd 5 10x 2 2.

y 5 22 72. False, e is an irrational number.e Þ271,80199,990 .

73.

Thus, but f sxd 5 hsxd.f sxd Þ gsxd,

5 hsxd

51

9s3xd

5 3x 1 1

322 5 3x322

f sxd 5 3x22 74.

Thus, but gsxd Þ fsxd.gsxd 5 hsxd

5 hsxd

5 64s4xd

5 64s22dx

5 64s22xd

5 22x ? 26

gsxd 5 22x16

75. and

Thus, f sxd 5 gsxd 5 hsxd.

5 gsxd

5 s14dx22

5 hsxd 5 s14d2s22xd

5 16s222xd 5 422x

5 16s22d2x 5 42s42xd

f sxd 5 16s42xd f sxd 5 16s42xd 76.

Thus, none are equal.

hsxd 5 25x23 5 2s5x ? 523d

gsxd 5 532x 5 53 ? 52x

fsxd 5 52x 1 3

(b)x Sample Data Model

0 12 12.5

25 44 44.5

50 81 81.82

75 96 96.19

100 99 99.3

70. (a)

Altitude (in km)

Atm

osp

her

ic p

ress

ure

(in p

asca

ls)

P

h5 10 15 20

40,000

60,000

80,000

25

100,000

120,000

20,000

(b)

5 32,357 pascals

5 107,428e20.150s8d

p 5 107,428e20.150h

(c) When

(d) when

x < 38 masses.

2

3s100d 5

100

1 1 7e20.069x

y 5100

1 1 7e20.069s36d < 63.14%.

x 5 36:


78. (a)

Decreasing:

Increasing:

Relative maximum:

Relative minimum: s0, 0d

s2, 4e22d

s0, 2d

s2`, 0d, s2, `d

−2 7

−1

5

f sxd 5 x2e2x (b)

Decreasing:

Increasing:

Relative maximum: s1.44, 4.25d

s2`, 1.44d

s1.44, `d

−2

−2

10

6

gsxd 5 x232x

81.

y 5 ±!25 2 x2

y2 5 25 2 x2

x2 1 y2 5 25 82.

y 5 x 2 2 and y 5 2sx 2 2d, x ≥ 2

x 2 2 5 |y|x 2 |y| 5 2

79. (Horizontal line)

As

As x → 2`, fsxd → gsxd.

x → `, fsxd → gsxd.

−3 3

0

g

f

4

f sxd 5 11 10.5

x 2x

and gsxd 5 e0.5 80. The functions (c) and (d) are exponential.22x3x

77.

(a)

(b) 4x> 3x when x > 0.

4x< 3x when x < 0.

x21−1−2

−1

1

2

3

y

y = 4x y = 3x

y 5 3x and y 5 4x

x 0 1 2

1 3 9

1 4 1614

1164x

13

193x

2122

83.

Vertical asymptote:

Horizontal asymptote: y 5 0

x 5 29

x3−3−15−18 −6

9

12

6

3

−9

−3

−6

yf sxd 52

9 1 x

x

2 12221f sxd

2728210211

Section 3.2 Logarithmic Functions and Their Graphs 273

Section 3.2 Logarithmic Functions and Their Graphs

n You should know that a function of the form where and is called

a logarithm of x to base a.

n You should be able to convert from logarithmic form to exponential form and vice versa.

n You should know the following properties of logarithms.

(a)

(b)

(c)

(d) Inverse Property

(e) If

n You should know the definition of the natural logarithmic function.

n You should know the properties of the natural logarithmic function.

(a)

(b)

(c)

(d) Inverse Property

(e) If

n You should be able to graph logarithmic functions.

ln x 5 ln y, then x 5 y.

eln x 5 x

ln ex 5 x since ex 5 ex .

ln e 5 1 since e1 5 e.

ln 1 5 0 since e0 5 1.

loge x 5 ln x, x > 0

loga x 5 loga y, then x 5 y.

aloga x 5 x

loga ax 5 x since ax 5 ax .

loga a 5 1 since a1 5 a.

loga 1 5 0 since a0 5 1.

y 5 loga x ⇔ ay 5 x

x > 0,a > 0, a Þ 1,y 5 loga x,

Vocabulary Check

1. logarithmic 2. 10 3. natural;

4. 5. x 5 yaloga x 5 x

e

1. log4 64 5 3 ⇒ 43 5 64 2. log3 81 5 4 ⇒ 34 5 81 3. log7 1

49 5 22 ⇒ 722 51

49

4. log 1

1000 5 23 ⇒ 1023 51

1000 5. log32 4 525 ⇒ 322y5 5 4 6. log16 8 5

34 ⇒ 163y4 5 8

7. log36 6 512 ⇒ 361y2 5 6 8. log8 4 5

23 ⇒ 82y3 5 4 9. 53 5 125 ⇒ log5 125 5 3

10. 82 5 64 ⇒ log8 64 5 2 11. 811y4 5 3 ⇒ log81 3 514 12. 93y2 5 27 ⇒ log9 27 5

32

85. Answers will vary.84.

Domain: s2`, 7g

y

x64 8

−6

−4

−2

2

4

6

2−2−4

f sxd 5 !7 2 x

3 6 7

4 3 2 1 0y

2229x


22.

since

b23 5 b23

gsb23d 5 logb b23 5 23

gsxd 5 logb x 23.

f s45d 5 logs4

5d < 20.097

f sxd 5 log x 24.

f s 1500d 5 log

1500 < 22.699

f sxd 5 log x

25.

f s12.5d < 1.097

f sxd 5 log x 26.

f s75.25d < 1.877

f sxd 5 log x 27. since 34 5 34log3 34 5 4

28.

Since log1.5 1 5 0.1.50 5 1,

log1.5 1 29. since p1 5 p.logp p 5 1 30.

Since 9log9 15 5 15.aloga x 5 x,

9log9 15

31.

Domain: The domain is

x-intercept:

Vertical asymptote:

y 5 log4 x ⇒ 4 y 5 x

x 5 0

s1, 0d

s0, `d.x > 0 ⇒

f sxd 5 log4 x

x 1 4 2

0 11221f sxd

14

−1 1 2 3

−2

−1

1

2

x

y

32.

Domain:

x-intercept:

Vertical asymptote:

y 5 log6 x ⇒ 6y 5 x

x 5 0

s1, 0d

s0, `d

1−1 2 3

−2

−1

1

2

x

ygsxd 5 log6 x

x 1 6

y 0 11221

!616

33.

Domain:

x-intercept:

The x-intercept is

Vertical asymptote:

log3 x 5 2 2 y ⇒ 322y 5 x

y 5 2log3 x 1 2

x 5 0

s9, 0d.

9 5 x

32 5 x

2 5 log3 x

2log3 x 1 2 5 0

s0, `d

y 5 2log3 x 1 2

x 27 9 3 1

y 0 1 2 321

13

2 4 6 8 10 12

−6

−4

−2

2

4

6

x

y 34.

Domain:

The domain is

x-intercept:

The x-intercept is

Vertical asymptote:

y 5 log4sx 2 3d ⇒ 4y 1 3 5 x

x 2 3 5 0 ⇒ x 5 3

s4, 0d.

4 5 x

1 5 x 2 3

40 5 x 2 3

log4sx 2 3d 5 0

2 4 6 8 10

−4

−2

2

4

6

x

ys3, `d.

x 2 3 > 0 ⇒ x > 3

hsxd 5 log4sx 2 3d

x 3 4 7 19

y 0 1 221

14

13. 622 5136 ⇒ log6

136 5 22 14. 423 5

164 ⇒ log4

164 5 23 15. 70 5 1 ⇒ log7 1 5 0

16. 1023 5 0.001 ⇒ log10 0.001 5 23 17.

f s16d 5 log2 16 5 4 since 24 5 16

f sxd 5 log2 x 18.

since 161y2 5 4fs4d 5 log16 4 512

fsxd 5 log16 x

19.

f s1d 5 log7 1 5 0 since 70 5 1

f sxd 5 log7 x 20.

since 101 5 10f s10d 5 log 10 5 1

fsxd 5 log x 21.

5 2 by the Inverse Property

gsa2d 5 loga a2

gsxd 5 loga x


35.

Domain:

The domain is

x-intercept:

The x-intercept is

Vertical asymptote:

62y 2 2 5 x

2y 5 log6sx 1 2d

2 y 5 2log6sx 1 2d

x 1 2 5 0 ⇒ x 5 22

s21, 0d.

21 5 x

1 5 x 1 2

60 5 x 1 2

0 5 log6sx 1 2d

0 5 2log6sx 1 2d

6

−4

−2

2

4

x

ys22, `d.

x 1 2 > 0 ⇒ x > 22

f sxd 5 2log6sx 1 2d

x 4

0 1 221f sxd

21353621

5621

36.

Domain:

The domain is

x-intercept:

The x-intercept is

Vertical asymptote:

y 5 log5sx 2 1d 1 4 ⇒ 5y24 1 1 5 x

x 2 1 5 0 ⇒ x 5 1

s626625, 0d.

626625 5 x

1

625 5 x 2 1

524 5 x 2 1

log5sx 2 1d 5 24

log5sx 2 1d 1 4 5 0

2 3 4 5 6

1

2

3

4

5

6

x

ys1, `d.

x 2 1 > 0 ⇒ x > 1

y 5 log5sx 2 1d 1 4

x 1.00032 1.0016 1.008 1.04 1.2

y 21 0 1 2 3

37.

Domain:

The domain is

x-intercept:

The x-intercept is

Vertical asymptote:

The vertical asymptote is the y-axis.

x

55 0 ⇒ x 5 0

s5, 0d.

x

55 1 ⇒ x 5 5

x

55 100

log1x

52 5 0

4 6 8

−4

−2

2

4

x

ys0, `d.

x

5> 0 ⇒ x > 0

y 5 log1x

52 x 1 2 3 4 5 6 7

y 0 0.150.0820.1020.2220.4020.70

38.

Domain:

The domain is

x-intercept:

The x-intercept is

Vertical asymptote:

y 5 logs2xd ⇒ 210y 5 x

x 5 0

s21, 0d.

21 5 x

100 5 2x

−3 −2 −1 1

−2

−1

1

2

x

ylogs2xd 5 0

s2`, 0d.

2x > 0 ⇒ x < 0

y 5 logs2xdx

y 0 12122

2102121

1021

100


39.

Asymptote:

Point on graph:

Matches graph (c).

The graph of is obtained by shifting the graph of

upward two units.

gsxdf sxd

s1, 2d

x 5 0

f sxd 5 log3 x 1 2 40.

Asymptote:

Point on graph:

Matches graph (f).

reflects in the axis.x-gsxdf sxd

s1, 0d

x 5 0

fsxd 5 2log3 x

41.

Asymptote:

Point on graph:

Matches graph (d).

The graph of is obtained by reflecting the graph of

about the axis and shifting the graph two units to

the left.

x-gsxdf sxd

s21, 0d

x 5 22

f sxd 5 2log3sx 1 2d 42.

Asymptote:

Point on graph:

Matches graph (e).

shifts one unit to the right.gsxdf sxd

s2, 0d

x 5 1

fsxd 5 log3sx 2 1d

43.

Asymptote:

Point on graph:

Matches graph (b).

The graph of is obtained by reflecting the graph of

about the axis and shifting the graph one unit to

the right.

y-gsxdf sxd

s0, 0d

x 5 1

f sxd 5 log3s1 2 xd 5 log3f2sx 2 1dg 44.

Asymptote:

Point on graph:

Matches graph (a).

reflects in the axis then reflects that graph in

the axis.y-

x-gsxdf sxd

s21, 0d

x 5 0

fsxd 5 2log3s2xd

45. ln 12 5 20.693 . . . ⇒ e20.693 . . . 5

12 46. ln

25 5 20.916 . . . ⇒ e20.916 . . . 5

25

47. ln 4 5 1.386 . . . ⇒ e1.386 . . . 5 4 48. ln 10 5 2.302 . . . ⇒ e2.302 . . . 5 10

49. ln 250 5 5.521 . . . ⇒ e5.521 . . . 5 250 50. ln 679 5 6.520 . . . ⇒ e6.520 . . . 5 679

51. ln 1 5 0 ⇒ e0 5 1 52. ln e 5 1 ⇒ e1 5 e

53. e3 5 20.0855 . . . ⇒ ln 20.0855 . . . 5 3 54. e2 5 7.3890 . . . ⇒ ln 7.3890 . . . 5 2

55. e1/251.6487 . . . ⇒ ln 1.6487 . . . 512 56. e1y3 5 1.3956 . . . ⇒ ln 1.3956 . . . 5

13

57. e20.5 5 0.6065 . . . ⇒ ln 0.6065 . . . 5 20.5 58. e24.1 5 0.0165 . . . ⇒ ln 0.0165 . . . 5 24.1

59. ex 5 4 ⇒ ln 4 5 x 60. e2x 5 3 ⇒ ln 3 5 2x

61.

f s18.42d 5 ln 18.42 < 2.913

f sxd 5 ln x 62.

fs0.32d 5 3 ln 0.32 < 23.418

fsxd 5 3 ln x

63.

gs0.75d 5 2 ln 0.75 < 20.575

gsxd 5 2 ln x 64.

gs12d 5 2ln

12 < 0.693

gsxd 5 2ln x


65.

by the Inverse Propertygse3d 5 ln e3 5 3

gsxd 5 ln x 66.

gse22d 5 ln e22 5 22

gsxd 5 ln x

67.

by the Inverse Propertygse22y3d 5 ln e22y3 5 223

gsxd 5 ln x 68.

gse25y2d 5 ln e25y2 5 252

gsxd 5 ln x

69.

Domain:

The domain is

intercept:

The intercept is

Vertical asymptote: x 2 1 5 0 ⇒ x 5 1

s2, 0d.x-

2 5 x

e0 5 x 2 1

0 5 lnsx 2 1d

x-

y

x32 4 5

−3

−2

−1

1

2

3

1−1

s1, `d.

x 2 1 > 0 ⇒ x > 1

f sxd 5 lnsx 2 1d

1.5 2 3 4

0 0.69 1.1020.69f sxd

x

70.

Domain:

The domain is

x-intercept:

The x-intercept is

Vertical asymptote:

y 5 lnsx 1 1d ⇒ ey 2 1 5 x

x 1 1 5 0 ⇒ x 5 21

s0, 0d.

0 5 x

1 5 x 1 1

e0 5 x 1 1

lnsx 1 1d 5 0

2−2 4 6 8x

6

4

2

ys21, `d.

x 1 1 > 0 ⇒ x > 21

hsxd 5 lnsx 1 1d

x 0 1.72 6.39 19.09

y 0 1 2 3212

20.39

71.

Domain:

The domain is

x-intercept:

The x-intercept is

Vertical asymptote: 2x 5 0 ⇒ x 5 0

s21, 0d.

21 5 x

e0 5 2x

0 5 lns2xd

s2`, 0d.

−3 −2 −1 1

−2

1

2

x

y2x > 0 ⇒ x < 0

gsxd 5 lns2xd

x

0 1.100.6920.69gsxd

23222120.5

72.

Domain:

The domain is

x-intercept:

The x-intercept is

Vertical asymptote:

y 5 lns3 2 xd ⇒ 3 2 ey 5 x

3 2 x 5 0 ⇒ x 5 3

s2, 0d.

2 5 x

1 5 3 2 x

e0 5 3 2 x

lns3 2 xd 5 0

−2 −1 1 2 4

−3

−2

−1

2

3

x

ys2`, 3d.

3 2 x > 0 ⇒ x < 3

fsxd 5 lns3 2 xd

x 2.95 2.86 2.63 2 0.28

y 23 22 21 0 1

73.

−1

−2

5

2

y1 5 logsx 1 1d 74.

−1 5

−2

2

f sxd 5 logsx 2 1d 75.

0

−3

9

3

y1 5 lnsx 2 1d


76.

−4 5

−3

3

f sxd 5 lnsx 1 2d 77.

0

−1

9

5

y 5 ln x 1 2 78.

−5 10

−6

4

f sxd 5 3 ln x 2 1

79.

x 5 3

x 1 1 5 4

log2sx 1 1d 5 log2 4 80.

x 5 12

x 2 3 5 9

log2sx 2 3d 5 log2 9

81.

x 5 7

2x 1 1 5 15

logs2x 1 1d 5 log 15 82.

x 595

5x 5 9

5x 1 3 5 12

logs5x 1 3d 5 log 12

83.

x 5 4

x 1 2 5 6

lnsx 1 2d 5 ln 6 84.

x 5 6

x 2 4 5 2

lnsx 2 4d 5 ln 2

85.

x 5 ±5

x2 5 25

x2 2 2 5 23

lnsx2 2 2d 5 ln 23 86.

x 5 22 or x 5 3

sx 2 3dsx 1 2d 5 0

x2 2 x 2 6 5 0

x2 2 x 5 6

lnsx2 2 xd 5 ln 6

87.

(a) When

When

(b) Total amounts:

(c) Interest charges:

(d) The vertical asymptote is The closer the pay-

ment is to $1000 per month, the longer the length of the

mortgage will be. Also, the monthly payment must be

greater than $1000.

x 5 1000.

301,123.20 2 150,000 5 $151,123.20

396,234 2 150,000 5 $246,234

s1254.68ds12ds20d 5 $301,123.20

s1100.65ds12ds30d 5 $396,234.00

t 5 12.542 ln1 1254.68

1254.68 2 10002 < 20 years

x 5 $1254.68:

t 5 12.542 ln1 1100.65

1100.65 2 10002 < 30 years

x 5 $1100.65:

t 5 12.542 ln1 x

x 2 10002, x > 1000 88.

(a)

The number of years required to multiply the original

investment by increases with However, the larger

the value of the fewer the years required to increase

the value of the investment by an additional multiple

of the original investment.

(b)

2 4 6 8 10 12

5

10

15

20

25

K

t

K,

K.K

t 5ln K

0.095

K 1 2 4 6 8 10 12

t 0 7.3 14.6 18.9 21.9 24.2 26.2


98.

(a)

fsxd 5ln x

x

x 1 5 10

0 0.00001380.000920.0460.2300.322f sxd

106104102

90.

(a)

(b)

(c) No, the difference is due to the logarithmic

relationship between intensity and number of

decibels.

b 5 10 log1 1022

102122 5 10 logs1010d 5 100 decibels

b 5 10 log1 1

102122 5 10 logs1012d 5 120 decibels

b 5 10 log1 I

10212289.

(a)

(b)

(c)

(d) f s10d 5 80 2 17 log 11 < 62.3

f s4d 5 80 2 17 log 5 < 68.1

f s0d 5 80 2 17 log 1 5 80.0

0

0 12

100

f std 5 80 2 17 logst 1 1d, 0 ≤ t ≤ 12

91. False. Reflecting about the line will determine

the graph of f sxd.y 5 xgsxd 92. True, log3 27 5 3 ⇒ 33 5 27.

93.

f and g are inverses. Their graphs

are reflected about the line y 5 x.

−2 −1 1 2

−2

−1

1

2

x

g

f

y

f sxd 5 3x, gsxd 5 log3 x 94.

and are inverses. Their graphs


gf

−2 −1 1 2

−2

−1

1

2

x

g

f

y

f sxd 5 5x, gsxd 5 log5 x 95 .

f and g are inverses. Their graphs


−2 −1 1 2

−2

−1

1

2

x

g

f

y

fsxd 5 ex, gsxd 5 ln x

96.

and are inverses. Their graphs are reflected

about the line y 5 x.

gf

−2 −1 1 2

−2

−1

1

2

x

g

f

y

f sxd 5 10x, gsxd 5 log10 x 97. (a)

The natural log function

grows at a slower rate

than the square root

function.

(b)

The natural log function

grows at a slower rate than

the fourth root function.0

0

20,000

g

f

15gsxd 5 4!xfsxd 5 ln x,

0

0

1000

f

g

40gsxd 5 !xfsxd 5 ln x,

(b) As

(c)

0

0

100

0.5

x → `, fsxd → 0 .


99. (a) False. If were an exponential function of then

but not 0. Because one point is

is not an exponential function of

(c) True.

For

y 5 3, 23 5 8

y 5 1, 21 5 2

y 5 0, 20 5 1

a 5 2, x 5 2y.

x 5 ay

x.ys1, 0d,a1 5 a,y 5 ax,

x,y (b) True.

For

(d) False. If were a linear function of the slope

between and and the slope between

and would be the same. However,

and

Therefore, is not a linear function of x.y

m2 53 2 1

8 2 25

2

65

1

3.m1 5

1 2 0

2 2 15 1

s8, 3ds2, 1ds2, 1ds1, 0d

x,y

x 5 8, log2 8 5 3

x 5 2, log2 2 5 1

x 5 1, log2 1 5 0

a 5 2, y 5 log2 x.

y 5 loga x

100. so, for example, if there is no value of y for which If then every power

of a is equal to 1, so x could only be 1. So, is defined only for and a > 1.0 < a < 1loga x

a 5 1,s22dy 5 24.a 5 22,y 5 loga x ⇒ ay 5 x,

101.

(a) (b) Increasing on

Decreasing on

(c) Relative minimum:

s1, 0d

s0, 1ds1, `d

−1

−2

8

4

fsxd 5 |ln x| 102. (a)

−9

−4

9

8

hsxd 5 lnsx2 1 1d

For Exercises 103–108, use and gxxc 5 x32 1.f xxc 5 3x 1 2

103.

5 15

5 8 1 7

5 f3s2d 1 2g 1 fs2d3 2 1g

s f 1 gds2d 5 f s2d 1 gs2d 104.

Therefore,

5 1.

5 23 1 1 1 3

s f 2 gds21d 5 3s21d 2 s21d3 1 3

5 3x 2 x3 1 3

5 3x 1 2 2 x3 1 1

f sxd 2 gsxd 5 3x 1 2 2 sx3 2 1d

105.

5 4300

5 s20ds215d

5 f3s6d 1 2gfs6d3 2 1g

s fgds6d 5 f s6dgs6d 106.

Therefore, 1 f

g 2s0d 53 ? 0 1 2

03 2 15 22.

f sxdgsxd 5

3x 1 2

x3 2 1

107.

5 1028

5 3s342d 1 2

5 f s342d

5 f ss7d3 2 1d

s f 8 gds7d 5 f sgs7dd 108.

Therefore,

5 273 2 1 5 2344.

sg 8 f ds23d 5 s3 ? s23d 1 2d3 2 1

sg 8 f d(xd 5 gs f sxdd 5 gs3x 1 2d 5 s3x 1 2d3 2 1

(b) Increasing on

Decreasing on

(c) Relative minimum:

s0, 0d

s2`, 0ds0, `d

Section 3.3 Properties of Logarithms 281

12. log1y4 5 5log 5

logs1y4d 5ln 5

lns1y4d< 21.161 13. log9s0.4d 5

log 0.4

log 95

ln 0.4

ln 9< 20.417

14. log20 0.125 5log 0.125

log 205

ln 0.125

ln 20< 20.694 15. log15 1250 5

log 1250

log 155

ln 1250

ln 15< 2.633

16. log3 0.015 5log 0.015

log 35

ln 0.015

ln 3< 23.823 17. log4 8 5

log2 8

log2 45

log2 23

log2 22

53

2

Section 3.3 Properties of Logarithms

n You should know the following properties of logarithms.

(a)

(b)

(c)

(d)

n You should be able to rewrite logarithmic expressions using these properties.

ln un 5 n ln uloga un 5 n loga u

ln1u

v2 5 ln u 2 ln vloga1u

v2 5 loga u 2 loga v

lnsuvd 5 ln u 1 ln vlogasuvd 5 loga u 1 loga v

loga x 5ln x

ln aloga x 5

log10 x

log10 aloga x 5

logb x

logb a

Vocabulary Check

1. change-of-base 2.

3. 4.

This is the Product Property. Matches (c). This is the Power Property. Matches (a).

5.

This is the Quotient Property. Matches (b).

loga u

v5 loga u 2 loga v

ln un 5 n ln ulogasuvd 5 loga u 1 loga v

log x

log a5

ln x

ln a

1. (a)

(b) log5 x 5ln x

ln 5

log5 x 5log x

log 52. (a)

(b) log3 x 5ln x

ln 3

log3 x 5log x

log 33. (a)

(b) log1y5 x 5ln x

lns1y5d

log1y5 x 5log x

logs1y5d

4. (a)

(b) log1y3 x 5ln x

lns1y3d

log1y3 x 5log x

logs1y3d5. (a)

(b) logx 3

105

lns3y10dln x

logx 3

105

logs3y10dlog x

6. (a)

(b) logx 3

45

lns3y4dln x

logx 3

45

logs3y4dlog x

7. (a)

(b) log2.6 x 5ln x

ln 2.6

log2.6 x 5log x

log 2.68. (a)

(b) log7.1 x 5ln x

ln 7.1

log7.1 x 5log x

log 7.19. log3 7 5

log 7

log 35

ln 7

ln 3< 1.771

10. log7 4 5log 4

log 75

ln 4

ln 7< 0.712 11. log1y2 4 5

log 4

logs1y2d5

ln 4

lns1y2d5 22.000


18.

5 4 1 4 log2 3

5 4 log2 2 1 4 log2 3

5 2 log2 22 1 4 log2 3

5 2 log2 4 1 4 log2 3

log2s42 ? 34d 5 log2 42 1 log2 3

4 19.

5 23 2 log5 2

5 log5 523 1 log5 2

21

5 log5 1

125 1 log5 12

log5 1

250 5 log5s 1125 ? 1

2d 20.

5 log 3 2 2

5 log 3 2 2 log 10

5 log 3 2 log 102

5 log 3 2 log 100

log 9

300 5 log 3

100

21.

5 6 1 ln 5

5 ln 5 1 6

lns5e6d 5 ln 5 1 ln e6 22.

5 ln 6 2 2

5 ln 6 2 2 ln e

ln 6

e25 ln 6 2 ln e2 23. log3 9 5 2 log3 3 5 2

24. log5 1

125 5 log5 523 5 23 log5 5 5 23s1d 5 23 25. log2

4!8 514 log2 2

3 534 log2 2 5

34s1d 5

34

26. log6 3!6 5 log6 61y3 5

13 log6 6 5

13s1d 5

13 27. log4 161.2 5 1.2slog4 16d 5 1.2 log4 4

2 5 1.2s2d 5 2.4

28.

5 20.2s4d 5 20.8

5 20.2 log3 34

log3 8120.2 5 20.2 log3 81 29. is undefined. is not in the domain of log3 x.29log3s29d

30. is undefined because

is not in the domain of

log2 x.

216

log2s216d 31. ln e4.5 5 4.5 32.

5 12

5 12s1d

3 ln e4 5 s3ds4d ln e

33.

5 21

2

5 0 21

2s1d

5 0 21

2 ln e

ln 1

!e5 ln 1 2 ln!e 34.

53

4

53

4s1d

53

4 ln e

ln 4!e3 5 ln e3y4 35. ln e2 1 ln e5 5 2 1 5 5 7

36.

5 7

5 ln e7

5 ln e12

e5

2 ln e6 2 ln e5 5 ln e12 2 ln e5

37.

5 2

5 2 log5 5

5 log5 52

5 log5 25

log5 75 2 log5 3 5 log5 75

3

38.

5 3

512s1d 1

52s1d

512 log4 4 1

52 log4 4

log4 2 1 log4 32 5 log4 41y2 1 log4 4

5y2 39. log4 5x 5 log4 5 1 log4 x


55.

5 4 ln x 11

2 ln y 2 5 ln z

5 ln x4 1 ln !y 2 ln z5

ln1x4!y

z5 2 5 ln x4!y 2 ln z5 56.

51

2 log2 x 1 4 log2 y 2 4 log2 z

5 log2 !x 1 log2 y4 2 log2 z

4

log2 !x y4

z45 log2 !x y4 2 log2 z

4

53.

51

3 ln x 2

1

3 ln y

51

3fln x 2 ln yg

ln 3!x

y5

1

3 ln

x

y54.

5 ln x 23

2 ln y

51

2s2 ln x 2 3 ln yd

51

2sln x2 2 ln y3d

ln!x2

y35 ln1x2

y321y2

51

2 ln1x2

y32

57.

5 2 log5 x 2 2 log5 y 2 3 log5 z

5 log5 x2 2 slog5 y

2 1 log5 z3d

log5 1 x2

y2z32 5 log5 x2 2 log5 y

2z3 58.

5 log x 1 4 log y 2 5 log z

5 log x 1 log y4 2 log z5

log xy4

z55 log xy4 2 log z5

59.

534 ln x 1

14

lnsx2 1 3d

514f3 ln x 1 lnsx2 1 3dg

514fln x3 1 lnsx2 1 3dg

ln 4!x3sx2 1 3d 514 ln x3sx2 1 3d 60.

5 ln x 112 lnsx 1 2d

5 ln x 1 lnsx 1 2d1y2

5 lnfxsx 1 2d1y2g

ln!x2sx 1 2d 5 lnfx2sx 1 2dg1y2

40. log3 10z 5 log3 10 1 log3 z 41. log8 x4 5 4 log8 x 42. log

y

25 log y 2 log 2

43.

5 1 2 log5 x

log5 5

x5 log5 5 2 log5 x 44. log6 z

23 5 23 log6 z 45. ln!z 5 ln z1y2 51

2 ln z

49.

5 ln z 1 2 lnsz 2 1d, z > 1

ln zsz 2 1d2 5 ln z 1 lnsz 2 1d2 50.

5 lnsx 1 1d 1 lnsx 2 1d 2 3 ln x

5 lnfsx 1 1dsx 2 1dg 2 ln x3

ln1x2 2 1

x3 2 5 lnsx2 2 1d 2 ln x3

46. ln 3!t 5 ln t1y3 513 ln t 47.

5 ln x 1 ln y 1 2 ln z

ln xyz2 5 ln x 1 ln y 1 ln z2 48.

5 log 4 1 2 log x 1 log y

log 4x2y 5 log 4 1 log x2 1 log y

51.

51

2 log2sa 2 1d 2 2 log2 3, a > 1

51

2 log2sa 2 1d 2 log2 3

2

log2 !a 2 1

95 log2

!a 2 1 2 log2 9 52.

5 ln 6 21

2 lnsx2 1 1d

5 ln 6 2 lnsx2 1 1d1y2

ln 6

!x2 1 15 ln 6 2 ln!x2 1 1


61. ln x 1 ln 3 5 ln 3x 62. ln y 1 ln t 5 ln yt 5 ln ty 63. log4 z 2 log4 y 5 log4 z

y

64. log5 8 2 log5 t 5 log5 8

t65. 2 log2sx 1 4d 5 log2sx 1 4d2 66.

2

3 log7sz 2 2d 5 log7sz 2 2d2y3

67. 1

4 log3 5x 5 log3s5xd1y4 5 log3

4!5x 68. 24 log6 2x 5 log6s2xd24 5 log6 1

16x4

69.

5 ln x

sx 1 1d3

ln x 2 3 lnsx 1 1d 5 ln x 2 lnsx 1 1d3 70.

5 ln 64sz 2 4d5

5 ln 64 1 lnsz 2 4d5

2 ln 8 1 5 lnsz 2 4d 5 ln 82 1 lnsz 2 4d5

71.

5 log x

y21 log z

3 5 log xz3

y2

log x 2 2 log y 1 3 log z 5 log x 2 log y2 1 log z3 72.

5 log3 x3y4

z4

5 log3 x3y4 2 log3 z

4

3 log3 x 1 4 log3 y 2 4 log3 z 5 log3 x3 1 log3 y

4 2 log3 z4

73.

5 ln x

sx2 2 4d4

5 ln x 2 lnsx2 2 4d4

5 ln x 2 4 lnsx2 2 4d

ln x 2 4flnsx 1 2d 1 lnsx 2 2dg 5 ln x 2 4 lnsx 1 2dsx 2 2d

74.

5 ln z4sz 1 5d4

sz 2 5d2

5 lnfzsz 1 5dg4 2 lnsz 2 5d2

4fln z 1 lnsz 1 5dg 2 2 lnsz 2 5d 5 4fln zsz 1 5dg 2 lnsz 2 5d2

75.

5 ln 3!xsx 1 3d2

x2 2 1

51

3 ln

xsx 1 3d2

x2 2 1

51

3fln xsx 1 3d2 2 lnsx2 2 1dg

1

3 f2 lnsx 1 3d 1 ln x 2 lnsx2 2 1dg 5

1

3 flnsx 1 3d2 1 ln x 2 lnsx2 2 1dg

76.

5 ln1 x3

x2 2 122

5 2 ln x3

x2 2 1

5 2fln x3 2 flnsx 1 1dsx 2 1dgg

5 2fln x3 2 flnsx 1 1d 1 lnsx 2 1dgg

2f3 ln x 2 lnsx 1 1d 2 lnsx 2 1dg 5 2fln x3 2 lnsx 1 1d 2 lnsx 2 1dg


80.

51

21 log7

!10 by Property 1 and Property 3

51

21

1

2 log7 10

51

2f1 1 log7 10g

log7!70 5

1

2 log7 70 5

1

2flog7 7 1 log7 10g 81.

When

5 60 decibels

5 120 1 10s26d

b 5 120 1 10 log 1026

I 5 1026:

5 120 1 10 log I

5 10flog I 1 12g

5 10flog I 2 log 10212g

b 5 10 log1 I

102122

82.

5 24 dB

5 10slogs250.79dd

5 10slogs2.5079 3 102dd

5 101log13.16 3 107

1.26 3 10522 5 10slogs3.16 3 107d 2 logs1.26 3 105dd

Difference 5 10 log13.16 3 1025

10212 2 2 10 log11.26 3 1027

10212 2

b 5 10 log1 I

102122 83.

With both stereos playing, the music is

decibels louder.

10 log 2 < 3

5 s120 1 10 log I d 1 10 log 2

5 120 1 10slog 2 1 log I d

b 5 120 1 10 logs2I d

77.

5 log 81 3!y s y 1 4d2

y 2 1 2 5 log8

3!y s y 1 4d2 2 log8s y 2 1d

51

3 log8 ys y 1 4d2 2 log8s y 2 1d

1

3 flog8 y 1 2 log8s y 1 4dg 2 log8s y 2 1d 5

1

3 flog8 y 1 log8s y 1 4d2g 2 log8s y 2 1d

78.

5 log4fx6sx 2 1d!x 1 1g 5 log4f!x 1 1sx 2 1dg 1 log4 x

6

512flog4sx 1 1dsx 2 1d2g 1 log4 x

6

12flog4sx 1 1d 1 2 log4sx 2 1dg 1 6 log4 x 5

12flog4sx 1 1d 1 log4sx 2 1d2g 1 log4 x

6

79.

The second and third expressions are equal by Property 2.

log2 32

45 log2 32 2 log2 4 Þ

log2 32

log2 4


84.

(a)

(b)

(c)

(d)

(e)

120

95

70

f s12d 5 90 2 15 ? logs12 1 1d 5 73.3

f s4d 5 90 2 15 ? logs4 1 1d 5 79.5

f s0d 5 90

f std 5 90 2 logst 1 1d15

f std 5 90 2 15 logst 1 1d, 0 ≤ t ≤ 12

85. By using the regression feature on a graphing calculator we obtain y < 256.24 2 20.8 ln x.

(f ) The average score will be 75 when months. See

graph in (e).

(g)

t 5 9 months

101 5 t 1 1

1 5 logst 1 1d

215 5 215 logst 1 1d

75 5 90 2 15 logst 1 1d

t 5 9

(c)

This graph is identical to in (b).T

T 5 e20.037t14 1 21

lnsT 2 21d 5 20.037t 1 4

0

0

30

5

86. (a)

(b)

See graph in (a).

(d)

(e) Since the scatter plot of the original data is so

nicely exponential, there is no need to do the

transformations unless one desires to deal with

smaller numbers. The transformations did not

make the problem simpler.

Taking logs of temperatures led to a linear scatter

plot because the log function increases very slowly

as the values increase. Taking the reciprocals of

the temperatures led to a linear scatter plot because

of the asymptotic nature of the reciprocal function.

x-

0

0

30

80

0

0

30

0.07

T 51

0.0012t 1 0.0161 21

1

T 2 215 0.0012t 1 0.016

T 5 54.4s0.964d t 1 21

T 2 21 5 54.4s0.964dt

0

0

30

80

(in minutes)

0 78 57 4.043 0.0175

5 66 45 3.807 0.0222

10 57.5 36.5 3.597 0.0274

15 51.2 30.2 3.408 0.0331

20 46.3 25.3 3.231 0.0395

25 42.5 21.5 3.068 0.0465

30 39.6 18.6 2.923 0.0538

1ysT 2 21dlnsT 2 21dT 2 21 s8CdT s8Cdt

87.

False, since 0 is not in the domain of

f s1d 5 ln 1 5 0

f sxd.f s0d Þ 0

f sxd 5 ln x 88.

True, because fsaxd 5 ln ax 5 ln a 1 ln x 5 f sad 1 f sxd.

x > 0a > 0,fsaxd 5 fsad 1 fsxd,


94. Let then and

logb un 5 logb b

nx 5 nx 5 n logb u

un 5 bnx.u 5 bxx 5 logb u,

95.

−3

−3

6

3

f sxd 5 log2 x 5log x

log 25

ln x

ln 296.

−1 5

−2

2

f sxd 5 log4 x 5log x

log 45

ln x

ln 497.

−3

−3

6

3

5log x

logs1y2d 5ln x

lns1y2d

f sxd 5 log1y2 x

98.

−1 5

−2

2

5log x

logs1y4d 5ln x

lns1y4d

f sxd 5 log1y4 x 99.

−1

−2

5

2

5log x

log 11.85

ln x

ln 11.8

f sxd 5 log11.8 x 100.

−1 5

−2

2

5log x

log 12.45

ln x

ln 12.4

f sxd 5 log12.4 x

101.

by Property 2f sxd 5 hsxd

1 2 3 4

−2

−1

1

2

x

f = hg

y

f sxd 5 ln x

2, gsxd 5

ln x

ln 2, hsxd 5 ln x 2 ln 2

93. Let and then

Then logbsuyvd 5 logbsbx2yd 5 x 2 y 5 logb u 2 logb v.

u

v5

bx

by5 bx2y

bx 5 u and by 5 v.y 5 logb v,x 5 logb u

89. False. fsxd 2 fs2d 5 ln x 2 ln 2 5 ln x

2Þ lnsx 2 2d 90. false

can’t be simplified further.

f s!xd 5 ln!x 5 ln x1y2 51

2 ln x 5

1

2 fsxd

!f sxd 5 !ln x

!f sxd 51

2 fsxd;

91. False.

fsud 5 2fsvd ⇒ ln u 5 2 ln v ⇒ ln u 5 ln v2 ⇒ u 5 v2

92. If then

True

0 < x < 1.f sxd < 0,


102.

ln 20 5 lns5 ? 22d 5 ln 5 1 ln 22 5 ln 5 1 2 ln 2 < 1.6094 1 2s0.6931d 5 2.9956

ln 18 5 lns32 ? 2d 5 ln 32 1 ln 2 5 2 ln 3 1 ln 2 < 2s1.0986d 1 0.6931 5 2.8903

ln 16 5 ln 24 5 4 ln 2 < 4s0.6931d 5 2.7724

ln 15 5 lns5 ? 3d 5 ln 5 1 ln 3 < 1.6094 1 1.0986 5 2.7080

ln 12 5 lns22 ? 3d 5 ln 22 1 ln 3 5 2 ln 2 1 ln 3 < 2s0.6931d 1 1.0986 5 2.4848

ln 10 5 lns5 ? 2d 5 ln 5 1 ln 2 < 1.6094 1 0.6931 5 2.3025

ln 9 5 ln 32 5 2 ln 3 < 2s1.0986d 5 2.1972

ln 8 5 ln 23 5 3 ln 2 < 3s0.6931d 5 2.0793

ln 6 5 lns2 ? 3d 5 ln 2 1 ln 3 < 0.6931 1 1.0986 5 1.7917

ln 5 < 1.6094

ln 4 5 lns2 ? 2d 5 ln 2 1 ln 2 < 0.6931 1 0.6931 5 1.3862

ln 3 < 1.0986

ln 2 < 0.6931

ln 2 < 0.6931, ln 3 < 1.0986, ln 5 < 1.6094

103.24xy22

16x23y5

24xx3

16yy25

3x4

2y3, x Þ 0 104. 12x2

3y 223

5 1 3y

2x223

5s3yd3

s2x2d35

27y3

8x6

105. s18x3y4d23s18x3y4d3 5s18x3y4d3

s18x3y4d35 1 if x Þ 0, y Þ 0. 106.

5xy

s y 1 xdyxy5

sxyd2

x 1 y

5xy

s1yxd 1 s1yyd

xysx21 1 y21d21 5xy

x21 1 y21

107.

x 1 1 5 0 ⇒ x 5 21

3x 2 1 5 0 ⇒ x 513

s3x 2 1dsx 1 1d 5 0

3x2 1 2x 2 1 5 0 108.

The zeros are x 514, 1.

x 2 1 5 0 ⇒ x 5 1

4x 2 1 5 0 ⇒ x 514

s4x 2 1dsx 2 1d 5 0

4x2 2 5x 1 1 5 0

109.

521 ± !97

6

x 521 ± !12 2 4s3ds28d

2s3d

3x2 1 x 2 8 5 0

s3x 1 1dsxd 5 s2ds4d

2

3x 1 15

x

4 110.

The zeros are 1 ±!31

2.

1 ±!31

25 x

2 ±!124

45 x

2s22d ±!s22d2 2 4s2ds215d

2s2d 5 x

0 5 2x2 2 2x 2 15

15 5 2x2 2 2x

5s3d 5 2xsx 2 1d

5

x 2 15

2x

3

Section 3.4 Exponential and Logarithmic Equations 289

Section 3.4 Exponential and Logarithmic Equations

Vocabulary Check

1. solve 2. (a) (b) 3. extraneous

(c) (d) xx

x 5 yx 5 y

1.

(a)

Yes, is a solution.

(b)

No, is not a solution.x 5 2

42s2d27 5 423 5164 Þ 64

x 5 2

x 5 5

42s5d27 5 43 5 64

x 5 5

42x27 5 64 2.

(a)

No, is not a solution.

(b)

No, is not a solution.x 5 2

23s2d11 5 27 5 128

x 5 2

x 5 21

23s21d11 5 222 514

x 5 21

23x11 5 32

3.

(a)


(b)

Yes, is a solution.

(c)

Yes, is a solution.x < 1.219

3e1.21912 5 3e3.219 < 75

x < 1.219

x 5 22 1 ln 25

3es221 ln 25d12 5 3eln 25 5 3s25d 5 75

x 5 22 1 ln 25

x 5 22 1 e25

3es221e25d12 5 3ee25Þ 75

x 5 22 1 e25

3ex12 5 75 4.

(a)

Yes, is a solution.

(b)


(c)

Yes, is an approximate solution.x 5 20.0416

2e5s20.0416d12 5 2e1.792 < 2s6.00144d < 12

x 5 20.0416

x 5ln 6

5 ln 2

< 2 ? 97.9995 5 195.999

< 2e2.58512

2e5[ln 6ys5 ln 2dg12 5 2esln 6yln 2d12

x 5ln 6

5 ln 2

x 51

5s22 1 ln 6d

5 2eln 6 5 2 ? 6 5 12

2e5fs1y5ds221 ln 6dg12 5 2e221 ln 612

x 51

5s22 1 ln 6d

2e5x12 5 12

n To solve an exponential equation, isolate the exponential expression, then take the logarithm of both sides.

Then solve for the variable.

1. 2.

n To solve a logarithmic equation, rewrite it in exponential form. Then solve for the variable.

1. 2.

n If and we have the following:

1.

2.

n Check for extraneous solutions.

ax 5 ay ⇔ x 5 y

loga x 5 loga y ⇔ x 5 y

a Þ 1a > 0

eln x 5 xaloga x 5 x

ln ex 5 xloga ax 5 x


5.

(a)

Yes, 21.333 is an approximate solution.

(b)


(c)

Yes, is a solution.x 5643

3s643 d 5 64

x 5643

x 5 24

3s24d 5 212 Þ 64

x 5 24

3s21.333d < 64

x < 21.333

log4s3xd 5 3 ⇒ 3x 5 43 ⇒ 3x 5 64 6.

(a)

Since is a solution.

(b)

Since is not a solution.

(c)

Since is not a solution.102 2 3210 Þ 100,

log2s97 1 3d 5 log2s100d

x 5 102 2 3 5 97

x 5 17210 Þ 20,

log2s17 1 3d 5 log2s20d

x 5 17

x 5 1021210 5 1024,

log2s1021 1 3d 5 log2s1024d

x 5 1021

log2sx 1 3d 5 10

7.

(a)


(b)

Yes, is a solution.

(c)

Yes, is an approximate solution.x < 163.650

lnf2s163.650d 1 3g 5 ln 330.3 < 5.8

x < 163.650

x 512s23 1 e5.8d

lnf2s12ds23 1 e5.8d 1 3g 5 lnse5.8d 5 5.8

x 512s23 1 e5.8d

x 512s23 1 ln 5.8d

lnf2s12ds23 1 ln 5.8d 1 3g 5 lnsln 5.8d Þ 5.8

x 512s23 1 ln 5.8d

lns2x 1 3d 5 5.8 8.

(a)

Yes, is a solution.

(b)

Yes, is an approximate solution.

(c)

No, is not a solution.x 5 1 1 ln 3.8

lns1 1 ln 3.8 2 1d 5 lnsln 3.8d < 0.289

x 5 1 1 ln 3.8

x < 45.701

lns45.701 2 1d 5 lns44.701d < 3.8

x < 45.701

x 5 1 1 e3.8

lns1 1 e3.8 2 1d 5 ln e3.8 5 3.8

x 5 1 1 e3.8

lnsx 2 1d 5 3.8

9.

x 5 2

4x 5 42

4x 5 16 10.

x 5 5

3x 5 35

3x 5 243 11.

x 5 25

2x 5 5

22x 5 25

s12dx

5 32 12.

x 5 23

2x 5 3

42x 5 43

s14dx

5 64

13.

x 5 2

ln x 5 ln 2

ln x 2 ln 2 5 0 14.

x 5 5

ln x 5 ln 5

ln x 2 ln 5 5 0 15.

x < 0.693

x 5 ln 2

ln ex 5 ln 2

ex 5 2 16.

x < 1.386

x 5 ln 4

ln ex 5 ln 4

ex 5 4

17.

x < 0.368

x 5 e21

eln x 5 e21

ln x 5 21 18.

x < 0.000912

x 5 e27

eln x 5 e27

ln x 5 27 19.

x 5 64

x 5 43

4log4 x 5 43

log4 x 5 3 20.

or 0.008 x 51

125

x 5 523

log5 x 5 23


28.

x 5 0, x 5 1

2xsx 2 1d 5 0

2x2 2 2x 5 0

2x2 5 x2 2 2x

e2x25 ex222x 29.

x < 1.465

x 5 log3 5 5log 5

log 3 or

ln 5

ln 3

log3 3x 5 log3 5

3x 5 5

4s3xd 5 20 30.

x < 1.723

x 5ln 16

ln 5

x 5 log516

5x 5 16

2s5xd 5 32

31.

x 5 ln 5 < 1.609

ln ex 5 ln 5

ex 5 5

2ex 5 10 32.

x 5 ln 914 < 3.125

ln ex 5 ln 914

ex 5914

4ex 5 91 33.

x 5 ln 28 < 3.332

ln ex 5 ln 28

ex 5 28

ex 2 9 5 19

34.

x < 2.015

x 5ln 37

ln 6

x 5 log6 37

6x 5 37

6x 1 10 5 47 35.

x 5ln 80

2 ln 3< 1.994

2x ln 3 5 ln 80

ln 32x 5 ln 80

32x 5 80 36.

x 5ln 3000

5 ln 6< 0.894

5x 5ln 3000

ln 6

s5xd ln 6 5 ln 3000

ln 65x 5 ln 3000

65x 5 3000

37.

t 5 2

2t

25 21

52ty2 5 521

52ty2 51

5

52ty2 5 0.20 38.

t 5 2ln 0.10

3 ln 4< 0.554

23t 5ln 0.10

ln 4

s23td ln 4 5 ln 0.10

ln 423t 5 ln 0.10

423t 5 0.10 39.

x 5 4

x 2 1 5 3

3x21 5 33

3x21 5 27

21.

Point of intersection:

s3, 8d

x 5 3

2x 5 23

2x 5 8

f sxd 5 gsxd 22.


s23, 9d

x 523

27x 5 272y3

27x 5 9

fsxd 5 gsxd 23.


s9, 2d

x 5 9

x 5 32

log3 x 5 2

f sxd 5 gsxd 24.

Point of intersection: s5, 0d

x 5 5

x 2 4 5 1

elnsx24d5e0

lnsx 2 4d 5 0

fsxd 5 gsxd

25.

x 5 21 or x 5 2

0 5 sx 1 1dsx 2 2d

0 5 x2 2 x 2 2

x 5 x2 2 2

e x 5 ex222 26.

x 5 22, x 5 4

sx 2 4dsx 1 2d 5 0

x2 2 2x 2 8 5 0

2x 5 x2 2 8

e2 x 5 ex228 27.

By the Quadratic Formula

or x < 20.618.x < 1.618

x2 2 x 2 1 5 0

x2 2 3 5 x 2 2

ex223 5 ex22

40.

x 5 8

x 2 3 5 5

x 2 3 5 log2 32

2x23 5 32


41.

5 3 2ln 565

ln 2< 26.142

x 53 ln 2 2 ln 565

ln 2

x ln 2 5 3 ln 2 2 ln 565

2x ln 2 5 ln 565 2 3 ln 2

3 ln 2 2 x ln 2 5 ln 565

s3 2 xd ln 2 5 ln 565

ln 232x 5 ln 565

232x 5 565 42.

x 52ln 431 2 ln 64

ln 8< 24.917

x ln 8 5 2ln 431 2 ln 64

2x ln 8 5 ln 431 1 ln 82

22 ln 8 2 x ln 8 5 ln 431

s22 2 xd ln 8 5 ln 431

ln 8222x 5 ln 431

8222x 5 431

43.

< 0.059

x 5 1

3 log13

22

3x 5 log13

22

log 103x 5 log13

22

103x 512

8

8s103xd 5 12 44.

< 6.146

x 5 6 1 log 7

5

x 2 6 5 log 7

5

log 10x26 5 log 7

5

10x26 57

5

5s10x26d 5 7 45.

x 5 1 1ln 7

ln 5< 2.209

x 2 1 5ln 7

ln 5

sx 2 1d ln 5 5 ln 7

ln 5x21 5 ln 7

5x21 5 7

3s5x21d 5 21

46.

x 5 6 2ln 5

ln 3< 4.535

2x 5ln 5

ln 32 6

6 2 x 5ln 5

ln 3

s6 2 xd ln 3 5 ln 5

ln 362x 5 ln 5

362x 5 5

8s362xd 5 40 47.

x 5ln 12

3< 0.828

3x 5 ln 12

e3x 5 12 48.

x 5ln 50

2< 1.956

2x 5 ln 50

ln e2x 5 ln 50

e2x 5 50

49.

5 ln 53 < 0.511

x 5 2ln 35

2x 5 ln 35

e2x 535

500e2x 5 300 50.

< 0.648

x 5 214 ln

340

24x 5 ln 340

ln e24x 5 ln 340

e24x 5340

1000e24x 5 75 51.

x 5 ln 1 5 0

ex 5 1

22ex 5 22

7 2 2ex 5 5 52.

< 2.120

x 5 ln 253

ln ex 5 ln 253

ex 5253

3ex 5 25

214 1 3ex 5 11


54.

x 5 3 2ln 3.5

2 ln 4< 2.548

22x 5 26 1ln 3.5

ln 4

6 2 2x 5ln 3.5

ln 4

6 2 2x 5 log4 3.5

4622x 5 3.5

8s4622xd 5 28

8s4622xd 1 13 5 41

55.

or

(No solution) x 5 ln 5 < 1.609

ex 5 5ex 5 21

sex 1 1dsex 2 5d 5 0

e2x 2 4ex 2 5 5 0 56.

x 5 ln 2 < 0.693 or x 5 ln 3 < 1.099

ex 5 2 or ex 5 3


e2x 2 5ex 1 6 5 0

57.

Not possible since for all x.

ex 2 450 ⇒ ex 5 4 ⇒ x 5 ln 4 < 1.386

ex> 0

ex 1 150 ⇒ ex 5 21


e2x 2 3ex 2 4 5 0 58.

Because the discriminant is there

is no solution.

92 2 4s1ds36d 5 263,

sexd219ex 1 36 5 0

e2x 1 9ex 1 36 5 0

53.

x 51

3 3logs8y3d

log 21 14 < 0.805

3x 2 15 log218

32 5logs8y3d

log 2 or

lns8y3dln 2

log2 23x21 5 log218

32

23x21 58

3

6s23x21d 5 16

6s23x21d 2 7 5 9

59.

x 5 2 ln 75 < 8.635

x

25 ln 75

exy2 5 75

25 5 100 2 exy2

500 5 20s100 2 exy2d

500

100 2 e xy25 20 60.

x 5 ln 7 < 1.946

2x 5 2ln 7

2x 5 ln 721

2x 5 ln 1

7

ln 1

75 ln e2x

1

75 e2x

8

72 1 5 e2x

8

75 1 1 e2x

400 5 350s1 1 e2xd

400

1 1 e2x5 350 61.

x 5ln 1498

2< 3.656

ln 1498 5 2x

1498 5 e2x

1500 5 2 1 e2x

3000 5 2s2 1 e2xd

3000

2 1 e2x5 2


63.

t 5ln 4

365 lns1 10.065365 d < 21.330

365t ln11 10.065

365 2 5 ln 4

ln11 10.065

365 2365t

5 ln 4

11 10.065

365 2365t

5 4

64.

t 5ln 21

9 ln 3.938225< 0.247

9t ln 3.938225 5 ln 21

ln 3.9382259t 5 ln 21

3.9382259t 5 21

14 22.471

40 29t

5 21 65.

t 5ln 2

12 lns1 10.1012 d < 6.960

12t ln11 10.10

12 2 5 ln 2

ln11 10.10

12 212t

5 ln 2

11 10.10

12 212t

5 2

66.

t 5ln 30

3 lns16 20.878

26 d < 0.409

3t ln116 20.878

26 2 5 ln 30

ln116 20.878

26 23t

5 ln 30

116 20.878

26 23t

5 30 67.

Algebraically:

The zero is x < 20.427.

x < 20.427

x 5 1 2 ln125

6 2

1 2 x 5 ln125

6 2

e12x 525

6

6e12x 5 25

−6 15

−30

6gsxd 5 6e12x 2 25

68.

The zero is 22.322.

22.322 < x

21 2 ln 3.75 5 x

1 1 ln 3.75 5 2x

ln 3.75 5 2x 2 1

3.75 5 e2x21

215 5 24e2x21

0 5 24e2x21 1 15−5 5

−20

20f sxd 5 24e2x21 1 15

62.

x 5ln 31

6< 0.572

ln 31 5 6x

ln 31 5 ln e6x

31 5 e6x

17 5 e6x 2 14

119 5 7se6x 2 14d

119

e6x 2 145 7

69.

Algebraically:

The zero is x < 3.847.

x < 3.847

x 52

3 ln1962

3 2

3x

25 ln1962

3 2

e3xy2 5962

3

3e3xy2 5 962

−6 9

−1200

300f sxd 5 3e3xy2 2 962


72.

The zero is 1.081.

−5 5

−7

13

x < 1.081

x 5ln 7

1.8

1.8x 5 ln 7

e1.8x 5 7

2e1.8x 5 27

2e1.8x 1 7 5 0

f sxd 5 2e1.8x 1 7 73.

Algebraically:

The zero is

−40 40

−10

2

t < 16.636.

t < 16.636

t 5 ln 8

0.125

0.125t 5 ln 8

e0.125t 5 8

e0.125t 2 850

hstd 5 e0.125t 2 8 74.

The zero is 1.236.

−5 5

−35

10

x < 1.236

x 5ln 29

2.724

2.724x 5 ln 29

e2.724x 5 29

f sxd 5 e2.724x 2 29

70.

The zero is 20.478.

x < 20.478

x 5 21.5 ln 1.375

22x

35 ln 1.375

e22xy3 5 1.375

8e22xy3 5 11 −3 7

−15

5 gsxd 5 8e22xy3 2 11 71.

Algebraically:

The zero is t < 12.207.

t < 12.207

t 5 ln 3

0.09

0.09t 5 ln 3

e0.09t 5 3 −20 40

−4

8gstd 5 e0.09t 2 3

75.

x 5 e23 < 0.050

ln x 5 23 76.

x 5 e2 < 7.389

eln x 5 e2

ln x 5 2 77.

x 5e2.4

2< 5.512

2x 5 e2.4

ln 2x 5 2.4 78.

x 5e

4< 0.680

4x 5 e

eln 4x 5 e1

ln 4x 5 1

79.

5 1,000,000.000

x 5 106

log x 5 6 80.

z 5100

3< 33.333

3z 5 100

10log 3z 5 102

log 3z 5 2 81.

x 5e10y3

5< 5.606

5x 5 e10y3

ln 5x 510

3

3 ln 5x 5 10 82.

x 5 e7y2 < 33.115

eln x 5 e7y2

ln x 57

2

2 ln x 5 7

83.

< 5.389

x 5 e2 2 2

x 1 2 5 e2

!x 1 2 5 e1

ln !x 1 2 5 1 84.

< 22,034.466

x 5 e10 1 8

x 2 8 5 e10

!x 2 8 5 e5

eln!x28 5 e5

ln!x 2 8 5 5 85.

< 0.513

x 5 e22y3

ln x 5 223

3 ln x 5 22

7 1 3 ln x 5 5 86.

< 0.264

x 5 e24y3

eln x 5 e24y3

ln x 5 243

26 ln x 5 8

2 2 6 ln x 5 10


87.

x 5 2s311y6d < 14.988

0.5x 5 311y6

3log3s0.5xd 5 311y6

log3s0.5xd 5116

6 log3s0.5xd 5 11 88.

x 5 1011y5 1 2 < 160.489

x 2 2 5 1011y5

10log10sx22d 5 1011y5

log10sx 2 2d 5115

5 log10sx 2 2d 5 11

89.

This negative value is extraneous. The equation has

no solution.

x 5e2

1 2 e2< 21.157

xs1 2 e2d 5 e2

x 2 e2x 5 e2

x 5 e2x 1 e2

x 5 e2sx 1 1d

x

x 1 15 e2

ln1 x

x 1 12 5 2

ln x 2 lnsx 1 1d 5 2 90.

The only solution is x 521 1 !1 1 4e

2< 1.223.

x 521 ± !1 1 4e

2

x2 1 x 2 e 5 0

xsx 1 1d 5 e1

elnfxsx11dg 5 e1

lnfxsx 1 1dg 5 1

ln x 1 lnsx 1 1d 5 1

91.

The negative value is extraneous. The only solution is

x 5 1 1 !1 1 e < 2.928.

5 1 ± !1 1 e 52 ± 2!1 1 e

2

x 52 ± !4 1 4e

2

x2 2 2x 2 e 5 0

xsx 2 2d 5 e1

lnfxsx 2 2dg 5 1

ln x 1 lnsx 2 2d 5 1 92.

The only solution is .x 523 1 !9 1 4e

2< 0.729

x 523 ± !9 1 4e

2

x2 1 3x 2 e 5 0

xsx 1 3d 5 e1

elnfxsx13dg 5 e1

lnfxsx 1 3dg 5 1

ln x 1 lnsx 1 3d 5 1

93.

Both of these solutions are extraneous, so the equation

has no solution.

x 5 22 or x 5 23

sx 1 2dsx 1 3d 5 0

x2 1 5x 1 6 5 0

x2 1 6x 1 5 5 x 2 1

sx 1 5dsx 1 1d 5 x 2 1

x 1 5 5x 2 1

x 1 1

lnsx 1 5d 5 ln1x 2 1

x 1 12ln sx 1 5d 5 lnsx 2 1d 2 lnsx 1 1d 94.

(The negative apparent solution is extraneous.)

3.303 < x

3 ±!13

25 x

2s23d ±!s23d2 2 4s1ds21d

2s1d 5 x

0 5 x2 2 3x 2 1

x 1 1 5 x2 2 2x

x 1 1

x 2 25 x

ln1x 1 1

x 2 22 5 ln x

lnsx 1 1d 2 lnsx 2 2d 5 ln x

95.

x 5 7

2x 2 3 5 x 1 4

log2s2x 2 3d 5 log2sx 1 4d


100.

The value is extraneous. The only solution is

x 5 9.

x 5 21

x 5 9 or x 5 21

sx 2 9dsx 1 1d 5 0

x2 2 8x 2 9 5 0

x2 2 8x 5 9

3log3sx228xd 5 32

log3fxsx 2 8dg 5 2

log3 x 1 log3sx 2 8d 5 2

97.

Quadratic Formula

Choosing the positive value of x (the negative value is

extraneous), we have

x 521 1 !17

2< 1.562.

x 521 ± !17

2

0 5 x2 1 x 2 4

x 1 4 5 x2 1 2x

x 1 4

x5 x 1 2

log1x 1 4

x 2 5 logsx 1 2d

logsx 1 4d 2 log x 5 logsx 1 2d 98.

The value is extraneous. The only solution is

x 5 2.

x 5 23

x 5 23 or x 5 2

sx 1 3dsx 2 2d 5 0

x2 1 x 2 6 5 0

xsx 1 2d 5 x 1 6

log2fxsx 1 2dg 5 log2sx 1 6d

log2 x 1 log2sx 1 2d 5 log2sx 1 6d

99.

x 5 2

2x 5 22

x 5 2x 2 2

x 5 2sx 2 1d

x

x 2 15 41y2

4log4fxysx21dg 5 41y2

log41 x

x 2 12 51

2

log4 x 2 log4sx 2 1d 51

2

96.

The apparent solution is extraneous, because the domain of the logarithm function is positive numbers,

and and are negative. There is no solution.2s27d 1 127 2 6

x 5 27

27 5 x

x 2 6 5 2x 1 1

logsx 2 6d 5 logs2x 1 1d

101.

The only solution is x 525s29 1 5!33d

8< 180.384.

x < 0.866 (extraneous) or x < 180.384

525s29 ± 5!33d

85

725 ± !515,625

8 x 5

725 ± !7252 2 4s4ds625d2s4d

4x2 2 725x 1 625 5 0

4x2 2 100x 1 625 5 625x

s2x 2 25d2 5 s25!xd2

2x 2 25 5 25!x

2x 5 25s1 1 !xd 5 25 1 25!x

8x 5 100s1 1 !xd

8x

1 1 !x5 102

log 8x

1 1 !x5 2

log 8x 2 logs1 1 !xd 5 2


102.

The only solution is x 51225 1 125!73

2< 1146.500.

x < 78.500 sextraneousd or x < 1146.500

x 51225 ± 125!73

2

x 51225 ± !1,140,625

2

x 51225 ± !s21225d2 2 4s1ds90,000d

2

x2 2 1225x 1 90,000 5 0

x2 2 600x 1 90,000 5 625x

sx 2 300d2 5 s25!x d2

x 2 300 5 25!x

4x 2 1200 5 100!x

4x 5 1200 1 100!x

4x 5 100s12 1 !x d

4x

12 1 !x5 100

10logs4xy(121!x dd5 102

log1 4x

12 1 !x2 5 2

log 4x 2 logs12 1 !x d 5 2

103.

From the graph we have

when

Algebraically:

x 5ln 7

ln 2< 2.807

x ln 2 5 ln 7

ln 2x 5 ln 7

2x 5 7

y 5 7.x < 2.807

y2 5 2x

−8 10

−2

10y1 5 7 104.

The solution is x < 2.197.

2.197 < x

22 ln 1

35 x

ln 1

35 2

x

2

1

35 e2xy2

−2

−200

10

800 500 5 1500e2xy2

105.

From the graph we have

when

Algebraically:

x 5 e3 < 20.086

ln x 5 3

3 2 ln x 5 0

y 5 3.x < 20.086

y2 5 ln x

−5

−1

30

5y1 5 3 106.

The solution is x < 14.182.

x < 14.182

x 5 e2.5 1 2

x 2 2 5 e2.5

elnsx22d 5 e2.5

lnsx 2 2d 5 2.5

24 lnsx 2 2d 5 210

−5

−3

30

18 10 2 4 lnsx 2 2d 5 0


107. (a)

t < 8.2 years

ln 2

0.0855 t

ln 2 5 0.085t

2 5 e0.085t

5000 5 2500e0.085t

A 5 Pert (b)

t < 12.9 years

ln 3

0.0855 t

ln 3 5 0.085t

3 5 e0.085t

7500 5 2500e0.085t

A 5 Pert 108. (a)

t < 5.8 years

ln 2

0.125 t

ln 2 5 0.12t

ln 2 5 ln e0.12t

2 5 e0.12t

5000 5 2500e0.12t

A 5 Pert

r 5 0.12 (b)

t 5 9.2 years

ln 3

0.125 t

ln 3 5 0.12t

ln 3 5 ln e0.12t

3 5 e0.12t

7500 5 2500e0.12t

A 5 Pert

r 5 0.12

109.

(a)

x < 1426 units

0.004x 5 ln 300

300 5 e0.004x

350 5 500 2 0.5se0.004xd

p 5 350

p 5 500 2 0.5se0.004xd

(b)

x < 1498 units

0.004x 5 ln 400

400 5 e0.004x

300 5 500 2 0.5se0.004xd

p 5 300

110.

(a) When

x 5 2lns6y11d

0.002< 303 units

ln 6

115 20.002x

ln 6

115 ln e20.002x

6

115 e20.002x

0.48 5 0.88e20.002x

4 5 3.52 1 0.88e20.002x

4

4 1 e20.002x5 0.88

0.12 5 1 24

4 1 e20.002x

600 5 500011 24

4 1 e20.002x2p 5 $600:

p 5 500011 24

4 1 e20.002x2(b) When

x 5 2lns8y23d

0.002< 528 units

ln 8

235 20.002x

ln 8

235 ln e20.002x

8

235 e20.002x

0.32 5 0.92e20.002x

4 5 3.68 1 0.92e20.002x

4

4 1 e20.002x5 0.92

0.08 5 1 24

4 1 e20.002x

400 5 500011 24

4 1 e20.002x2p 5 $400:

111.

(a)

0

0

1500

10

V 5 6.7e248.1yt , t ≥ 0

(b) As

Horizontal asymptote:

The yield will approach

6.7 million cubic feet per acre.

V 5 6.7

t → `, V → 6.7. (c)

t 5248.1

lns13y67d< 29.3 years

ln113

672 5248.1

t

1.3

6.75 e248.1yt

1.3 5 6.7e248.1yt


112.

When

x 5 2log10s21y68d

0.04< 12.76 inches

log10 21

685 20.04x

21

685 1020.04x

21 5 68s1020.04xd

N 5 21:

N 5 68s1020.04xd 113.

corresponds to the year 2001.t < 11

t 5 e2.4 < 11

ln t 5 2.4

2630.0 ln t 5 21512

7312 2 630.0 ln t 5 5800

y 5 7312 2 630.0 ln t, 5 ≤ t ≤ 12

114.

Since represents 1995, indicates that the number of daily fee golf facilities in the U.S.

reached 9000 in 2001.

t 5 11.6t 5 5

t 5 e2.45222 5 11.6

ln t 54619

1883.65 2.45222

4619 5 1883.6 ln t

9000 5 4381 1 1883.6 ln t

y 5 4381 1 1883.6 ln t, 5 ≤ t ≤ 13

115. (a) From the graph shown in the textbook, we see horizontal asymptotes at and

These represent the lower and upper percent bounds; the range falls between 0% and 100%.

y 5 100.y 5 0

(b) Males

x 5 69.71 inches

20.6114sx 2 69.71d 5 0

20.6114sx 2 69.71d 5 ln 1

e20.6114sx269.71d 5 1

1 1 e20.6114sx269.71d 5 2

50 5100

1 1 e20.6114sx269.71d

Females

x 5 64.51 inches

20.66607sx 2 64.51d 5 0

20.66607sx 2 64.51d 5 ln 1

e20.6667sx264.51d 5 1

1 1 e20.66607sx264.51d 5 2

50 5100

1 1 e20.66607sx264.51d

(c) When

n 5 2

ln10.83

0.602 12

0.2< 5 trials

20.2n 5 ln10.83

0.602 12

ln e20.2n 5 ln10.83

0.602 12

e20.2n 50.83

0.602 1

1 1 e20.2n 50.83

0.60

0.60 50.83

1 1 e20.2n

P 5 60% or P 5 0.60:

116.

(a)

(b) Horizontal asymptotes:

The upper asymptote, indicates that the

proportion of correct responses will approach 0.83

as the number of trials increases.

P 5 0.83,

P 5 0, P 5 0.83

0

0 40

1.0

P 50.83

1 1 e20.2n


117.

(a)

(b)

The model seems to fit the data well.

(c) When

Add the graph of to the graph in part (a) and

estimate the point of intersection of the two graphs.

We find that meters.

(d) No, it is probably not practical to lower the number

of gs experienced during impact to less than 23

because the required distance traveled at is

meters. It is probably not practical to

design a car allowing a passenger to move forward

2.27 meters (or 7.45 feet) during an impact.

x < 2.27

y 5 23

x < 1.20

y 5 30

30 5 23.00 1 11.88 ln x 136.94

x

y 5 30:

0

0

1.2

200

y 5 23.00 1 11.88 ln x 136.94

x

x 0.2 0.4 0.6 0.8 1.0

y 162.6 78.5 52.5 40.5 33.9

119.

True by Property 1 in Section 3.3.

logasuvd 5 loga u 1 loga v 120.

False.

2.04 < log10s10 1 100d Þ slog10 10dslog10 100d 5 2

logasu 1 vd 5 sloga udsloga vd

121.

False.

Þ log 100 2 log 10 5 1

1.95 < logs100 2 10d

logasu 2 vd 5 loga u 2 loga v 122.

True by Property 2 in Section 3.3.

loga1u

v2 5 loga u 2 loga v 123. Yes, a logarithmic equation can

have more than one extraneous

solution. See Exercise 93.

124.

(a) This doubles your money.

(b)

(c)

Doubling the interest rate yields the same result as

doubling the number of years.

If (i.e., ), then doubling your

investment would yield the most money. If

then doubling either the interest rate

or the number of years would yield more money.

rt > ln 2,

rt < ln 22 > ert

A 5 Pers2td 5 Pertert 5 ertsPertd

A 5 Pes2rdt 5 Pertert 5 ertsPertd

A 5 s2Pdert 5 2sPertd

A 5 Pert 125. Yes.

Time to Double Time to Quadruple

Thus, the time to quadruple is twice as long as the

time to double.

2 ln 2

r5 t

ln 2

r5 t

ln 4 5 rt ln 2 5 rt

4 5 ert 2 5 ert

4P 5 Pert 2P 5 Pert

118.

(a) From the graph in the textbook we see a horizontal

asymptote at This represents the room

temperature.

(b)

h < 0.81 hour

lns4y7d2ln 2

5 h

ln14

72 5 2h ln 2

ln14

72 5 ln 22h

4

75 22h

4 5 7s22hd

5 5 1 1 7s22hd

100 5 20f1 1 7s22hdg

T 5 20.

T 5 20f1 1 7s22hdg


126. (a) When solving an exponential equation, rewrite the

original equation in a form that allows you to use the

One-to-One Property if and only if or

rewrite the original equation in logarithmic form and

use the Inverse Property loga ax 5 x.

x 5 yax 5 ay

(b) When solving a logarithmic equation, rewrite the

original equation in a form that allows you to use the

One-to-One Property if and only if

or rewrite the original equation in exponential

form and use the Inverse Property aloga x 5 x.

x 5 y

loga x 5 loga y

127.

5 4|x|y2!3y

!48x2y5 5 !16x2y43y 128.

5 4!2 2 10

!32 2 2!25 5 !16 ? 2 2 2s5d 129.

5 3!125 ? 3 5 5 3!3

3!25 3!15 5 3!375

130.

51

2!10 1 1

5!10 1 2

2

53s!10 1 2d

6

53s!10 1 2d

10 2 4

3

!10 2 25

3

!10 2 2?!10 1 2

!10 1 2131.

Domain: all real numbers

intercept:

axis symmetryy-

s0, 9dy-

x

y

x−2−4−6−8 2 4 6 8

−2

2

4

6

8

12

14

f sxd 5 |x| 1 9

0

9 10 11 12y

±3±2±1x

132. y

x−2−4−6 2 4 6 8

−2

−4

−6

2

4

6

8

133.

Domain: all real numbers

intercept:

intercept: s0, 4dy-

s2, 0dx-

x

y

x−1−2−3−4 1 3 4

−3

1

2

3

4

5

gsxd 5 52x,

2x2 1 4,

x < 0

x ≥ 0

0 1 2 3

4 3 2 2521222426y

20.5212223x

134. y

x64

−6

−2

1

4

6

2−2−4−6

135. log6 9 5log10 9

log10 65

ln 9

ln 6< 1.226 136. log3 4 5

log10 4

log10 35

ln 4

ln 3< 1.262

137. log3y4 5 5log10 5

log10s3y4d 5ln 5

lns3y4d < 25.595 138. log8 22 5log10 22

log10 85

ln 22

ln 8< 1.486

Section 3.5 Exponential and Logarithmic Models

Section 3.5 Exponential and Logarithmic Models 303

n You should be able to solve growth and decay problems.

(a) Exponential growth if

(b) Exponential decay if

n You should be able to use the Gaussian model

n You should be able to use the logistic growth model

n You should be able to use the logarithmic models

y 5 a 1 b log x.y 5 a 1 b ln x,

y 5a

1 1 be2rx.

y 5 ae2sx2bd2yc.

b > 0 and y 5 ae2bx.

b > 0 and y 5 aebx.

Vocabulary Check

1. 2. 3. normally distributed

4. bell; average value 5. sigmoidal

y 5 a 1 b ln x; y 5 a 1 b log xy 5 aebx; y 5 ae2bx

1.

This is an exponential growth

model. Matches graph (c).

y 5 2exy4 2.

This is an exponential decay

model. Matches graph (e).

y 5 6e2xy4 3.

This is a logarithmic function

shifted up six units and left two

units. Matches graph (b).

y 5 6 1 logsx 1 2d

4.

This is a Gaussian model.

Matches graph (a).

y 5 3e2sx22d2y5 5.

This is a logarithmic model shifted

left one unit. Matches graph (d).

y 5 lnsx 1 1d 6.

This is a logistic growth model.

Matches graph (f).

y 54

1 1 e22x

7. Since the time to double is given by

and we have

Amount after 10 years: A 5 1000e0.35 < $1419.07

t 5ln 2

0.035< 19.8 years.

ln 2 5 0.035t

ln 2 5 ln e0.035t

2 5 e0.035t

2000 5 1000e0.035t

A 5 1000e0.035t, 8. Since the time to double is given by

and we have

Amount after 10 years: A 5 750e0.105s10d < $2143.24

t 5ln 2

0.105< 6.60 years.

ln 2 5 0.105t

ln 2 5 ln e0.105t

2 5 e0.105t

1500 5 750e0.105t

1500 5 750e0.105t,

A 5 750e0.105t,


9. Since when we have

the following.

Amount after 10 years: A 5 750e0.089438s10d < $1834.37

r 5ln 2

7.75< 0.089438 5 8.9438%

ln 2 5 7.75r

ln 2 5 ln e7.75r

2 5 e7.75r

1500 5 750e7.75r

t 5 7.75,A 5 750ert and A 5 1500 10. Since and when

we have

Amount after 10 years:

A 5 10,000e0.057762s10d < $17,817.97

r 5ln 2

12< 0.057762 5 5.7762%.

ln 2 5 12r

ln 2 5 ln e12r

2 5 e12r

20,000 5 10,000e12r

t 5 12,A 5 20,000A 5 10,000ert

11. Since when

we have the following.

The time to double is given by

t 5ln 2

0.110< 6.3 years.

1000 5 500e0.110t

r 5lns1505.00y500d

10< 0.110 5 11.0%

1505.00 5 500e10r

t 5 10,

A 5 500ert and A 5 $1505.00 12. Since and when we have

The time to double is given by

t 5ln 2

0.3466< 2 years.

1200 5 600e0.3466t

r 5lns19,205y600d

10< 0.3466 or 34.66%.

ln119,205

600 2 5 10r

ln119,205

600 2 5 ln e10r

19,205

6005 e10r

19,205 5 600e10r

t 5 10,A 5 19,205A 5 600ert

13. Since and when

we have the following.

The time to double is given by t 5ln 2

0.045< 15.40 years.

10,000.00

e0.045s10d 5 P < $6376.28

10,000.00 5 Pe0.045s10d

t 5 10,A 5 10,000.00A 5 Pe0.045t 14. Since and when we have

The time to double is given by years.t 5ln 2

0.025 34.7

P 52000

e0.02s10d 5 $1637.46.

2000 5 Pe0.02s10d

t 5 10,A 5 2000A 5 Pe0.02t

15.

5500,000

1.00625240< $112,087.09

P 5500,000

11 10.075

12 212s20d

500,000 5 P11 10.075

12 212s20d16.

P 5 $4214.16

500,000 5 P11 10.12

12 212(40)

A 5 P11 1r

n2nt


17.

(a)

(c)

t 5ln 2

365 lns1 10.11365 d < 6.302 years

365t ln11 10.11

365 2 5 ln 2

11 10.11

365 2365t

5 2

n 5 365

t 5ln 2

ln 1.11< 6.642 years

t ln 1.11 5 ln 2

s1 1 0.11dt 5 2

n 5 1

P 5 1000, r 5 11%

(b)

(d) Compounded continuously

t 5ln 2

0.11< 6.301 years

0.11t 5 ln 2

e0.11t 5 2

t 5ln 2

12 lns1 10.1112 d < 6.330 years

12t ln11 10.11

12 2 5 ln 2

11 10.11

12 212t

5 2

n 5 12

18.

(a)

(c)

t 5ln 2

365 lns1 10.105365 d < 6.602 years

n 5 365

t 5ln 2

lns1 1 0.105d< 6.94 years

n 5 1

r 5 10.5% 5 0.105P 5 1000,

(b)

(d) Compounded continuously

t 5ln 2

0.105< 6.601 years

t 5ln 2

12 lns1 10.105

12 d < 6.63 years

n 5 12

19.

ln 3

r5 t

ln 3 5 rt

3 5 ert

3P 5 Pert

r 2% 4% 6% 8% 10% 12%

(years) 9.1610.9913.7318.3127.4754.93t 5ln 3

r

20.

Using the power regression feature of a graphing utility, t 5 1.099r21.

0

0

0.16

60

21.

ln 3

lns1 1 rd5 t

ln 3 5 t lns1 1 rd

ln 3 5 lns1 1 rdt

3 5 s1 1 rdt

3P 5 Ps1 1 rdt

r 2% 4% 6% 8% 10% 12%

(years) 55.48 28.01 18.85 14.27 11.53 9.69t 5ln 3

lns1 1 rd


22.

Using the power regression feature of a graphing utility,

t 5 1.222r21.

0

0

0.16

60 23. Continuous compounding results in faster growth.

Time (in years)

Am

ount

(in d

oll

ars)

t

A t= 1 + 0.075 [[ [[

A e= 0.07t

A

1.00

1.25

1.50

1.75

2.00

2 4 6 8 10

A 5 1 1 0.075v t b and A 5 e0.07t

24.

From the graph, compounded

daily grows faster than 6% simple

interest.

512%

0

2

A t= 1 + 0.06[[ [[

A = 1 +0.055365( ) 365t[[ [[

0 10

25.

Given grams after

1000 years, we have

< 6.48 grams.

y 5 10efsln 0.5dy1599gs1000d

C 5 10

k 5ln 0.5

1599

ln 0.5 5 ks1599d

ln 0.5 5 ln eks1599d

0.5 5 eks1599d

1

2C 5 Ceks1599d 26.

Given grams after 1000

years, we have

C < 2.31 grams.

1.5 5 Ceflns1y2dy1599gs1000d

y 5 1.5

k 5lns1y2d1599

ln 1

25 ks1599d

ln 1

25 ln eks1599d

1

25 eks1599d

1

2C 5 Ceks1599d

27.


years, we have

C < 2.26 grams.

2 5 Cefsln 0.5dy5715gs1000d

y 5 2

k 5ln 0.5

5715

ln 0.5 5 ks5715d

ln 0.5 5 ln eks5715d

0.5 5 eks5715d

1

2C 5 Ceks5715d 28.

Given grams, after 1000

years we have

y < 2.66 grams.

y 5 3eflns1y2dy5715gs1000d

C 5 3

k 5lns1y2d

5715

ln 1

25 ks5715d

ln 1

25 ln eks5715d

1

25 eks5715d

1

2C 5 Ceks5715d 29.


years, we have

C < 2.16 grams.

2.1 5 Cefsln 0.5dy24,100gs1000d

y 5 2.1

k 5ln 0.5

24,100

ln 0.5 5 ks24,100d

ln 0.5 5 ln eks24,100d

0.5 5 eks24,100d

1

2C 5 Ceks24,100d


36.

—CONTINUED—

30.


years, we have

C < 0.41 grams.

0.4 5 Ceflns1y2dy24,100gs1000d

y 5 0.4

k 5lns1y2d 24,100

ln 1

25 ks24,100d

ln 1

25 ln eks24,100d

1

25 eks24,100d

1

2C 5 Ceks24,100d 31.

Thus, y 5 e0.7675x .

ln 10

35 b ⇒ b < 0.7675

ln 10 5 3b

10 5 ebs3d

1 5 aebs0d ⇒ 1 5 a

y 5 aebx 32.

Thus, y 512e0.5756x.

ln 10

45 b ⇒ b < 0.5756

ln 10 5 4b

ln 10 5 ln e4b

10 5 e4b

5 51

2ebs4d

1

25 aebs0d ⇒ a 5

1

2

y 5 aebx

33.

Thus, y 5 5e20.4024x.

lns1y5d

45 b ⇒ b < 20.4024

ln11

52 5 4b

1

55 e4b

1 5 5ebs4d

5 5 aebs0d ⇒ 5 5 a

y 5 aebx 34.

Thus, y 5 e20.4621x .

lns1y4d

35 b ⇒ b < 20.4621

ln11

42 5 3b

ln11

42 5 ln e3b

1

45 ebs3d

1 5 aebs0d ⇒ 1 5 a

y 5 aebx

35.

(a) Since the exponent is negative, this is an exponential

decay model. The population is decreasing.

(b) For 2000, let thousand people

For 2003, let thousand peopleP < 2408.95t 5 3:

P 5 2430t 5 0:

P 5 2430e20.0029t

(c)

The population will reach 2.3 million (according to

the model) during the later part of the year 2018.

t 5lns2300y2430d

20.0029< 18.96

ln12300

24302 5 20.0029t

2300

24305 e20.0029t

2300 5 2430e20.0029t

2.3 million 5 2300 thousand

Country 2000 2010

Bulgaria 7.8 7.1

Canada 31.3 34.3

China 1268.9 1347.6

United Kingdom 59.5 61.2

United States 282.3 309.2


36. —CONTINUED—

(a) Bulgaria:

For 2030, use

million

China:

For 2030, use

million

United Kingdom:

For 2030, use

milliony 5 59.5e0.00282s30d < 64.7

t 5 30.

ln 61.2

59.55 10b ⇒ b < 0.00282

61.2 5 59.5ebs10d

a 5 59.5

y 5 1268.9e0.00602s30d < 1520.06

t 5 30.

ln 1347.6

1268.95 10b ⇒ b < 0.00602

1347.6 5 1268.9ebs10d

a 5 1268.9

y 5 7.8e20.0094s30d 5 5.88

t 5 30.

ln 7.1

7.85 10b ⇒ b 5 20.0094

7.1 5 7.8ebs10d

a 5 7.8

Canada:

For 2030, use

million

United States:

For 2030, use

milliony 5 282.3e0.0091s30d < 370.9

t 5 30.

ln 309.2

282.35 10b ⇒ b < 0.0091

309.2 5 282.3ebs10d

a 5 282.3

y 5 31.3e0.00915s30d < 41.2

t 5 30.

ln 34.3

31.35 10b ⇒ b < 0.00915

34.3 5 31.3ebs10d

a 5 31.3

(b) The constant b determines the growth rates. The greater the rate of growth, the greater the value of b.

(c) The constant b determines whether the population is increasing or decreasing .sb < 0dsb > 0d

37.

When

When hitsy 5 4080e0.2988s24d < 5,309,734t 5 24:

k 5lns10,000y4080d

3< 0.2988

ln110,000

4080 2 5 3k

10,000

40805 e3k

10,000 5 4080eks3d

y 5 10,000:t 5 3,

y 5 4080ekt 38.

For 2010,

milliony 5 10es0.1337ds20d 5 $144.98

t 5 20:

ln165

102 5 14k ⇒ k 5 0.1337

65 5 10eks14d

y 5 10ekt


39.

t 5ln 2

0.2197< 3.15 hours

200 5 100e0.2197t

N 5 100e0.2197t

k 5ln 3

5< 0.2197

ln 3 5 5k

ln 3 5 ln e5k

3 5 e5k

300 5 100e5k

N 5 100ekt 40.

hours t 5ln 2

sln 1.12dy10< 61.16

ln 2 5 1ln 1.12

10 2t

2 5 efsln 1.12dy10gt

500 5 250efsln 1.12dy10gt

N 5 250efsln 1.12dy10gt

k 5ln 1.12

10

1.12 5 e10k

280 5 250eks10d

N 5 250ekt

41.

(a)

(b)

t 5 28223 ln11012

13112 < 4797 years old

2t

82235 ln11012

13112

e2ty8223 51012

1311

1

1012e2ty8223 5

1

1311

t 5 28223 ln11012

814 2 < 12,180 years old

2t

82235 ln11012

814 2

e2ty8223 51012

814

1

1012e2ty8223 5

1

814

R 51

814

R 51

1012e2ty8223 42.

The ancient charcoal has only 15% as much radioactive

carbon.

years t 55715 ln 0.15

ln 0.5< 15,642

ln 0.15 5ln 0.5

5715t

0.15C 5 Cefsln 0.5dy5715gt

k 5lns1y2d5715

ln 1

25 5715k

1

2C 5 Ce5715k

y 5 Cekt

43.

(a)

Linear model:

—CONTINUED—

V 5 26394t 1 30,788

b 5 30,788

m 518,000 2 30,788

2 2 05 26394

s0, 30,788d, s2, 18,000d

(b)

Exponential model: V 5 30,788e20.268t

k 51

2 ln14500

76972 < 20.268

ln14500

76972 5 2k

4500

76975 e2k

18,000 5 30,788eks2d

a 5 30,788 (c)

The exponential model

depreciates faster in the

first two years.

0 4

0

32,000


43. —CONTINUED—

(d) (e) The linear model gives a higher value for the car for the

first two years, then the exponential model yields a higher

value. If the car is less than two years old, the seller would

most likely want to use the linear model and the buyer the

exponential model. If it is more than two years old, the

opposite is true.

1 3

$24,394 $11,606

$23,550 $13,779V 5 30,788e20.268t

V 5 26394t 1 30,788

t

44.

(a)

(c)

The exponential model depreciates faster in the

first two years.

(e) The slope of the linear model means that the comput-

er depreciates $300 per year, then loses all value in

the third year. The exponential model depreciates

faster in the first two years but maintains value longer.

0 4

0

1200

V 5 2300t 1 1150

m 5550 2 1150

2 2 05 2300

s0, 1150d, s2, 550d

(b)

(d)

V 5 1150e20.369t

ln1 550

11502 5 2k ⇒ k < 20.369

550 5 1150eks2d

t 1 3

$850 $250

$795 $380V 5 1150e20.369t

V 5 2300t 1 1100

45.

(a)

Sstd 5 100s1 2 e20.1625td

k < 20.1625

k 5 ln 0.85

ln 0.85 5 ln ek

0.85 5 ek

85100 5 ek

285 5 2100ek

15 5 100s1 2 eks1dd

Sstd 5 100s1 2 ektd

(b)

(c) Ss5d 5 100s1 2 e20.1625s5dd < 55.625 5 55,625 units

Time (in years)

Sal

es

(in t

housa

nds

of

unit

s)

S

t5 3025201510

30

60

90

120

46.

(a)

So, N 5 30s1 2 e20.050d.

k 5 20.050

20k 5 ln111

302

ln e20k 5 ln111

302

e20k 511

30

30e20k 5 11

19 5 30s1 2 e20kd

N 5 19, t 5 20

N 5 30s1 2 ektd

(b)

t 5lns5y30d20.050

5 36 days

ln1 5

302 5 20.050t

ln1 5

302 5 ln e20.050t

5

305 e20.050t

25 5 30s1 2 e20.050td

N 5 25


47.

(a)

(b) The average IQ score of an adult student is 100.

70 115

0

0.04

y 5 0.0266e2sx2100d2y450, 70 ≤ x ≤ 116 48. (a)

(b) The average number of hours per week a student uses

the tutor center is 5.4.

4 7

0

0.9

49.

(a) animals

(b)

months

(c)

The horizontal asymptotes are and

The asymptote with the larger -value,

indicates that the population size will approach 1000

as time increases.

p 5 1000,p

p 5 1000.p 5 0

0

0

40

1200

t 5 2lns1y9d0.1656

< 13

e20.1656t 51

9

9e20.1656t 5 1

1 1 9e20.1656t 5 2

500 51000

1 1 9e20.1656t

ps5d 51000

1 1 9e20.1656s5d < 203

pstd 51000

1 1 9e20.1656t50.

(a)

So,

(b) When

S 5500,000

1 1 0.6e0.0263s8d < 287,273 units sold.

t 5 8:

S 5500,000

1 1 0.6e0.0263t.

k 51

4 ln110

9 2 < 0.0263

4k 5 ln110

9 2

e4k 510

9

0.6e4k 52

3

1 1 0.6e4k 55

3

300,000 5500,000

1 1 0.6e4k

S 5500,000

1 1 0.6ekt

51. since

(a)

(b)

(c) 4.2 5 log I ⇒ I 5 104.2 < 15,849

8.3 5 log I ⇒ I 5 108.3 < 199,526,231

7.9 5 log I ⇒ I 5 107.9 < 79,432,823

I0 5 1.R 5 log I

I0

5 log I 52. since

(a)

(b)

(c) R 5 log 251,200 5 5.40

R 5 log 48,275,000 5 7.68

R 5 log 80,500,000 5 7.91

I0 5 1.R 5 log I

I0

5 log I

53.

(a)

(c) b 5 10 log 1028

102125 10 log 104 5 40 decibels

b 5 10 log 10210

102125 10 log 102 5 20 decibels

b 5 10 log I

I0

where I0 5 10212 wattym2.

(b)

(d) b 5 10 log 1

102125 10 log 1012 5 120 decibels

b 5 10 log 1025

102125 10 log 107 5 70 decibels


54. where

(a)

(c) bs1024d 5 10 log 1024

102125 10 log 108 5 80 decibels

bs10211d 5 10 log 10211

102125 10 log 101 5 10 decibels

I0 5 10212 wattym2bsId 5 10 log I

I0

(b)

(d) bs1022d 5 10 log 1022

102125 10 log 1010 5 100 decibels

bs102d 5 10 log 102

102125 10 log 1014 5 140 decibels

55.

% decrease 5I0109.3 2 I0108.0

I0109.33 100 < 95%

I 5 I010by10

10by10 5I

I0

10by10 5 10log IyI0

b

105 log

I

I0

b 5 10 log I

I0

56.

% decrease 5I0108.8 2 I0107.2

I0108.83 100 < 97%

I 5 I010by10

10by10 5I

I0

b 5 10 log10 I

I0

57.

2logs2.3 3 1025d < 4.64

pH 5 2logfH1g 58.

2logf11.3 3 1026g < 4.95

pH 5 2logfH1g

59.

fH1g < 1.58 3 1026 mole per liter

1025.8 5 fH1g

1025.8 5 10logfH1g

25.8 5 logfH1g

5.8 5 2logfH1g 60.

mole per liter fH1g < 6.3 3 1024

1023.2 5 fH1g

3.2 5 2logfH1g

61.

times the hydrogen ion

concentration of drinking water

1022.9

10285 105.1

fH1g 5 1028 for the drinking water

28.0 5 logfH1g

8.0 5 2logfH1g

fH1g 5 1022.9 for the apple juice

22.9 5 logfH1g

2.9 5 2logfH1g 62.

The hydrogen ion concentration is increased by

a factor of 10.

102pH ? 10 5 fH1g

102pH11 5 fH1g

102spH21d 5 fH1g

2spH 2 1d 5 logfH1g

pH 2 1 5 2logfH1g

63.

At 9:00 A.M. we have:

hours

From this you can conclude that the person died at 3:00 A.M.

t 5 210 ln 85.7 2 70

98.6 2 70< 6

t 5 210 ln T 2 70

98.6 2 70


64. Interest:

Principal:

(a)

(b) In the early years of the mortgage, the majority of the

monthly payment goes toward interest. The principal

and interest are nearly equal when t < 26 years.

0

0

35

v

u

800

P 5 120,000, t 5 35, r 5 0.075, M 5 809.39

v 5 1M 2Pr

12211 1r

12212t

u 5 M 2 1M 2Pr

12211 1r

12212t

(c)

The interest is still the majority of the monthly payment

in the early years. Now the principal and interest are

nearly equal when t < 10.729 < 11 years.

u

v

0

0

20

800

P 5 120,000, t 5 20, r 5 0.075, M 5 966.71

65.

(a)

0

240

150,000

u 5 120,0003 0.075t

1 2 1 1

1 1 0.075y12212t

2 14(b) From the graph, when years. It

would take approximately 37.6 years to pay $240,000 in

interest. Yes, it is possible to pay twice as much in interest

charges as the size of the mortgage. It is especially likely

when the interest rates are higher.

t < 21u 5 $120,000

66.

(a) Linear model:

Exponential model: or

(b)

20

0

100

t3

t2

t1

t4

25

t4 5 1.5385s1.0296ds

t4 5 1.5385e0.02913s

t3 5 0.2729s 2 6.0143

t2 5 1.2259 1 0.0023s2

t1 5 40.757 1 0.556s 2 15.817 ln s

(c)

Note: Table values will vary slightly depending on the

model used for t4.

s 30 40 50 60 70 80 90

3.6 4.6 6.7 9.4 12.5 15.9 19.6

3.3 4.9 7.0 9.5 12.5 15.9 19.9

2.2 4.9 7.6 10.4 13.1 15.8 18.5

3.7 4.9 6.6 8.8 11.8 15.8 21.2t4

t3

t2

t1

(d) Model :

Model :

Model :

Model :

The quadratic model, best fits the data.t2,

|15.8 2 15.9| 1 |20 2 21.2| 5 2.6

S4 5 |3.4 2 3.7| 1 |5 2 4.9| 1 |7 2 6.6| 1 |9.3 2 8.9| 1 |12 2 11.9| 1t4

|15.8 2 15.8| 1 |20 2 18.5| 5 5.6

S3 5 |3.4 2 2.2| 1 |5 2 4.9| 1 |7 2 7.6| 1 |9.3 2 10.4| 1 |12 2 13.1| 1t3

|15.8 2 15.9| 1 |20 2 19.9| 5 1.1

S2 5 |3.4 2 3.3| 1 |5 2 4.9| 1 |7 2 7| 1 |9.3 2 9.5| 1 |12 2 12.5| 1t2

|15.8 2 15.9| 1 |20 2 19.6| 5 2.0

S1 5 |3.4 2 3.6| 1 |5 2 4.6| 1 |7 2 6.7| 1 |9.3 2 9.4| 1 |12 2 12.5| 1t1


69. False. The graph of is the graph of shifted

upward five units.

gsxdf sxd 70. True. Powers of e are always positive, so if a

Gaussian model will always be greater than 0, and if

a Gaussian model will always be less than 0.a < 0,

a > 0,

67. False. The domain can be the set of real numbers for a

logistic growth function.

68. False. A logistic growth function never has an x-intercept.

71. (a) Logarithmic

(b) Logistic

(c) Exponential (decay)

(d) Linear

(e) None of the above (appears to be a combination

of a linear and a quadratic)

(f) Exponential (growth)

72. Answers will vary.

73.

(a)

(b)

(c) Midpoint:

(d) m 55 2 2

0 2 s21d 53

15 3

121 1 0

2,

2 1 5

2 2 5 121

2,

7

22d 5 !s0 2 s21dd2 1 s5 2 2d2 5 !12 1 32 5 !10

y

x−1−2−3 1 2 3

−1

1

2

3

5 (0, 5)

(−1, 2)

s21, 2d, s0, 5d 74.

(a)

(b)

(c) Midpoint:

(d) m 523 2 1

4 2 s26d 524

105 2

2

5

126 1 4

2,

23 1 1

2 2 5 s21, 21d

5 !100 1 16 5 !116 5 2!29

d 5 !s26 2 4d2 1 s1 2 s23dd2

y

x

(−6, 1)

(4, −3)

−4−6 2 4 6

−2

−4

−6

2

4

6

s4, 23d, s26, 1d

75.

(a)

(b)

(c) Midpoint:

(d) m 522 2 3

14 2 35 2

5

11

13 1 14

2,

3 1 s22d2 2 5 117

2,

1

22 5 !112 1 s25d2 5 !146

d 5 !s14 2 3d2 1 s22 2 3d2

y

x−2 2 4 6 8 10 14

−2

−4

−6

−8

2

4

6

8

(3, 3)

(14, −2)

s3, 3d, s14, 22d 76.

(a)

(b)

(c) Midpoint:

(d) m 54 2 0

10 2 75

4

3

17 1 10

2,

0 1 4

2 2 5 117

2, 22

5 !9 1 16 5 !25 5 5

d 5 !s10 2 7d2 1 s4 2 0d2

y

x64 8 10

−6

−4

−2

2

4

6

2−2

(7, 0)

(10, 4)

s10, 4d, s7, 0d


77.

(a)

(b)

(c) Midpoint:

(d) m 50 2 s21y4d

s3y4d 2 s1y2d 51y4

1y45 1

1s1y2d 1 s3y4d2

, s21y4d 1 0

2 2 5 15

8, 2

1

82

5!11

422

1 11

422

5!1

8

d 5!13

42

1

222

1 10 2 121

4222

y

x1

1

1

2

1

4( (, −

3

4( (, 0

1

2−

1

2

1

2

−

11

2, 2

1

42, 13

4, 02 78.

(a)

(b)

(c) Midpoint:

(d) m 5s21y3d 2 s1y6ds22y3d 2 s7y3d 5

21y2

235

1

6

1s22y3d 1 s7y3d2

, s21y3d 1 s1y6d

2 2 5 15

6, 2

1

122

5!s23d2 1 11

222

5 !9.25

d 5!122

32

7

322

1 121

32

1

622

y

x

2

3

1

3( (, −−

−1 1 2 3

−2

1

2

7

3

1

6( (,

17

3,

1

62, 122

3, 2

1

32

79.

Line

Slope:

y-intercept: s0, 10d

m 5 23

x

8

8

6

4

2

62−2−2

10 12

10

yy 5 10 2 3x 80.

Line

Slope:

y-intercept: s0, 21d

m 5 24

x

3

−2

−3

1−1−2−3

2

−1

32

yy 5 24x 2 1

81.

Parabola

Vertex: s0, 23d

y 5 22sx 2 0d2 2 3x

642

2

−2

−2−4−6

yy 5 22x2 2 3 82.

Parabola

Vertex:

x-intercepts: s252, 0d, s6, 0d

s74, 2

2898 d

5 2sx 274d2

2289

8

5 s2x 1 5dsx 2 6d x

−5

−30

−35

2 4−4 8

y y 5 2x2 2 7x 2 30

83.

Parabola

Vertex:

Focus:

Directrix: y 5 213

s0, 13d

s0, 0d

x2 543 y

3x2 5 4y

x

4321−1−2−3−4

7

6

5

4

3

2

1

y 3x2 2 4y 5 0 84.

Parabola

Vertex:

Focus:

Directrix: y 5 2

s0, 22d

s0, 0d

x2 5 28yx

−2

−4

−6

−10

64−4−6

2

−8

y 2x2 2 8y 5 0


85.

Vertical asymptote:


x

21−1

−1

−2

−3

−2−3

3

1

y

y 5 0

x 51

3

y 54

1 2 3x86.

Vertical asymptote:

Slant asymptote:

x

−4−6−8 4

2

4

6

8

10

y

y 5 2x 1 2

x 5 22

y 5x2

2x 2 25 2x 1 2 1

4

2x 2 2

87.

Circle

Center:

Radius: 5

s0, 8d

x

8642

14

12

10

8

6

4

2

−2−4−6−8

y x2 1 sy 2 8d2 5 25 88.

Parabola

Vertex:

Focus:

Directrix: y 5 22.75

s4, 23.25d

P 5 214

s4, 23d x

−4

4

−6

−8

2−2 6 8

−2

−10

y

sx 2 4d2 5 2s y 1 3d

sx 2 4d2 5 2y 2 7 1 4

sx 2 4d2 1 s y 1 7d 5 4

89.


x

8642−2−4−6 10

14

10

8

4

6

2

12

y

y 5 5

f sxd 5 2x21 1 5

x 25 23 21 0 1 3 5

5.02 5.06 5.3 5.5 6 9 21f sxd

90.


x

−4

−6

−10

2

−8

−2

y

y 5 21

f sxd 5 222x21 2 1

x 0 1 2

2982

542

322223f sxd

2122

91.


x

4

5

4

3

2

1

32−1−2

−2

−3

−5

−3−4−5−6

yf sxd 5 3x 2 4

x 24 22 21 0 1 2

23.99 23.89 23.67 23 21 5f sxd

Review Exercises for Chapter 3 317

92.


x

2

1

5

1 2 3 4 5−1−2

−2

−3

−4

−5

−3−4−5

yf sxd 5 23x1 4

x 0 1 2

3 1 253233

89f sxd

2122

Review Exercises for Chapter 3

1.

f s2.4d 5 6.12.4 < 76.699

f sxd 5 6.1x 2.

f s!3 d 5 30!35 361.784

f sxd 5 30x 3.

f spd 5 220.5spd < 0.337

f sxd 5 220.5x

4.

f s1d 5 12781y55 4.181

f sxd 5 1278xy5 5.

< 1456.529

f s2!11 d 5 7s0.22!11d f sxd 5 7s0.2xd 6.

f s20.8d 5 214s520.8d 5 23.863

f sxd 5 214s5xd

7.

Intercept:

Horizontal asymptote: x-axis

Increasing on:

Matches graph (c).

s2`, `d

s0, 1d

f sxd 5 4x 8.

Intercept:


Decreasing on:

Matches graph (d).

s2`, `d

y 5 0

s0,1d

fsxd 5 42x 9.

Intercept:

Horizontal asymptote: x-axis

Decreasing on:

Matches graph (a).

s2`, `d

s0, 21d

f sxd 5 24x

10.

Intercept:


Increasing on:

Matches graph (b).

s2`, `d

y 5 1

s0, 2d

fsxd 5 4x1 1 11.

Since the graph

of g can be obtained by shifting

the graph of f one unit to the right.

gsxd 5 f sx 2 1d,

gsxd 5 5x21

f sxd 5 5x 12.

Because the

graph of g can be obtained by

shifting the graph of f three units

downward.

gsxd 5 f sxd 2 3,

f sxd 5 4x, gsxd 5 4x2 3

13.

Since the graph of g can be obtained

by reflecting the graph of f about the x-axis and shifting

two units to the left.2f

gsxd 5 2f sx 1 2d,

gsxd 5 2s12dx12

f sxd 5 s12dx

14.


obtained by reflecting the graph of f in the x-axis and

shifting the graph of f eight units upward.

gsxd 5 2f sxd 1 8,

f sxd 5 s23dx

, gsxd 5 8 2 s23dx

15.


x

2

2

8

4−2−4

yf sxd 5 42x1 4

x 0 1 2 3

8 5 4.0164.0634.25f sxd

21

93. Answers will vary.


16.


x

321−1−2−3−4−5−6

1

−2

−3

−4

−5

−6

−7

−8

y

y 5 23

f sxd 5 24x2 3

x 22 21 0 1 2

23.063 23.25 24 27 219f sxd

17.


x

−3

−6

−9

−12

−15

63−3−6 9

y

y 5 0

f sxd 5 22.65x11

x 0 1 2

218.6127.02322.652120.377f sxd

2122

18.


x321−1−2−3

5

4

3

2

1

−1

y

y 5 0

f sxd 5 2.65x21

x 23 21 0 1 3

0.020 0.142 0.377 1 7.023f sxd

19.


x

6

2

2

8

4−2−4

y

y 5 4

f sxd 5 5x221 4

x 0 1 2 3

5 94.24.044.008f sxd

21

20.


x

6

4

2

−2

−4

−6

642−2 10

y

y 5 25

f sxd 5 2x262 5

x 0 5 6 7 8 9

24.984 24.5 24 23 21 3f sxd

21.


x

2

2

6

8

4−2−4

y

y 5 3

f sxd 5 s12d2x

1 3 5 2x1 3

x 0 1 2

4 5 73.53.25f sxd

2122


22.


x

4

2

−2

−4

−6

2−4

yf sxd 5 s18dx12

2 5

x 23 22 21 0 2

3 24 24.875 24.984 25f sxd

23.

x 5 24

x 1 2 5 22

3x125 322

3x125

19 24.

x 5 22

x 2 2 5 24

s13dx22

5 s13d24

s13dx22

5 34

s13dx22

5 81 25.

x 5225

5x 5 22

5x 2 7 5 15

e5x275 e15 26.

x 5112

22x 5 211

8 2 2x 5 23

e822x5 e23

27. e8 < 2980.958 28. e5y8 < 1.868 29. e21.7 < 0.183 30. e0.2785 1.320

31.

−4 −3 −2 −1 1 2 3 4

2

3

4

5

6

7

x

y

hsxd 5 e2xy2

x 0 1 2

2.72 1.65 1 0.61 0.37hsxd

2122

32.

−4 −3 −1 1 2 3 4

−5

−4

−3

−2

3

x

y

hsxd 5 2 2 e2xy2

x 22 21 0 1 2

y 20.72 0.35 1 1.39 1.63

33.

−6 −5 −4 −3 −2 −1 1 2

1

2

6

7

x

y

f sxd 5 ex12

x 0 1

1 20.097.392.720.37f sxd

212223

34.

1 2 3 4 5

1

2

3

4

5

t

y

sstd 5 4e22yt, t > 0

t 1 2 3 4

y 0.07 0.54 1.47 2.05 2.43

12


35. A 5 350011 10.065

n 210n

or A 5 3500es0.065ds10d


A $6704.39$6704.00$6692.64$6669.46$6635.43$6569.98

36. A 5 200011 10.05

n 230n

or A 5 2000es0.05ds30d

n 1 2 4 12 365 Continuous

A $8643.88 $8799.58 $8880.43 $8935.49 $8962.46 $8963.38

37.

(a) (b) (c) Fs5d < 0.811Fs2d < 0.487Fs 12d < 0.154

Fst) 5 1 2 e2ty3

38.

(a)

(b)

(c) According to the model, the car depreciates most

rapidly at the beginning. Yes, this is realistic.

Vs2d 5 14,000s34d2

5 $7875

0

0

10

15,000

Vstd 5 14,000s34d

t39. (a)

(b) The doubling time is ln 2

0.0875< 7.9 years.

A 5 50,000es0.0875ds35d < $1,069,047.14

40.

(a) For grams (b) For grams

(c)

Time (in years)

Mas

s of

24

1P

u (

in g

ram

s) 100

80

60

40

20

20 40 60 80 100

Q

t

t 5 10: Q 5 100s12d10y14.4 < 61.79t 5 0: Q 5 100s1

2d0y14.45 100

Q 5 100s12dty14.4

41.

log4 64 5 3

435 64 42.

log25 125 532

253y25 125 43.

ln 2.2255 . . . 5 0.8

e0.85 2.2255 . . .

44.

ln 1 5 0

e05 1 45.

5 log 1035 3

f s1000d 5 log 1000

f sxd 5 log x 46. log9 3 5 log9 91y2

512


47.

gs18d 5 log2s1

8d 5 log2 223

5 23

gsxd 5 log2 x 48.

f s14d 5 log4

14 5 21

f sxd 5 log4 x 49.

x 5 7

x 1 7 5 14

log4sx 1 7d 5 log4 14

50.

x 5 5

3x 5 15

3x 2 10 5 5

log8s3x 2 10d 5 log8 5 51.

x 5 25

x 1 9 5 4

lnsx 1 9d 5 ln 4 52.

x 5 6

2x 5 12

2x 2 1 5 11

lns2x 2 1d 5 lns11d

53.

Domain:

x-intercept:

Vertical asymptote: x 5 0

s1, 0d

s0, `d

x4321−1−2

4

3

2

1

−1

−2

ygsxd 5 log7 x ⇒ x 5 7y

x 1 7 49

0 1 221gsxd

17

54.

Domain:

x-intercept:


s1, 0d

x 5 1

x 5 50

log5 x 5 0

s0, `d

−1 1 2 3 4 5

−3

−2

−1

1

2

3

x

ygsxd 5 log5 x ⇒ 5y5 x

x 1 5 25

22 21 0 1 2gsxd

15

125

55.

Domain:

x-intercept:

Vertical asymptote:

x 5 0

s3, 0d

x

5432−1

3

2

1

−1

−2

−3

ys0, `d

f sxd 5 log1x

32 ⇒ x

35 10y ⇒ x 5 3s10yd

x 3 30

0 12122f sxd

0.30.03

56.

Domain:

x-intercept:


s0.000001, 0d

x 5 0.000001

x 5 1026

log x 5 26

6 1 log x 5 0

s0, `d

x8642−2

−2

10

8

10

6

4

2

yf sxd 5 6 1 log x

x 1 2 4 6 8 10

6 6.3 6.6 6.8 6.9 7f sxd

57.

Domain:

x-intercept:


x 5 1042 5 5 9995.

x 1 5 5 104

Since 4 2 logsx 1 5d 5 0 ⇒ logsx 1 5d 5 4

s9995, 0d

s25, `d

x21−1−2−3−4−6

7

6

5

4

3

2

1

yf sxd 5 4 2 logsx 1 5dx 0 1

4 3.223.303.403.523.70f sxd

21222324


58.

Domain:

x-intercept:


s3.1, 0d

x 5 3.1

x 2 3 5 1021

logsx 2 3d 5 21

logsx 2 3d 1 1 5 0

s3, `d

x98765

5

4

3

2

1

−2

−3

−4

−5

421−1

yf sxd 5 logsx 2 3d 1 1

x 4 5 6 7 8

1 1.3 1.5 1.6 1.7f sxd

59. ln 22.6 < 3.118 60. ln 0.98 < 20.020 61. ln e2125 212

62. ln e75 7 63. lns!7 1 5d < 2.034 64. ln1!3

8 2 < 21.530

65.

Domain:

x-intercept:


se23, 0d

x 5 e23

ln x 5 23

ln x 1 3 5 0

s0, `d

1−1 2 3 4 5

1

2

3

4

5

6

x

yf sxd 5 ln x 1 3

x 1 2 3

3 1.612.314.103.69f sxd

14

12

66.

Domain:

x-intercept:


s4, 0d

x 5 4

x 2 3 5 e0

lnsx 2 3d 5 0

s3, `d

2 4 6 8

−4

−2

2

4

x

yfsxd 5 lnsx 2 3d

x 3.5 4 4.5 5 5.5

y 20.69 0 0.41 0.69 0.92

67.

Domain:

x-intercepts:


s±1, 0d

s2`, 0d < s0,`d

x4321−1−2−3−4

4

3

2

1

−3

−4

yhsxd 5 lnsx2d 5 2 ln|x|

x

y 0 2.772.201.3921.39

±4±3±2±1±0.5

68.

Domain:

x-intercept:


s1, 0d

x 5 1

x 5 e0

ln x 5 0

14 ln x 5 0

s0, `d

1 2 3 4 5 6

−3

−2

−1

1

2

3

x

yfsxd 5

14 ln x

x 1 2 3

y 20.17 0 0.10 0.17 0.23 0.27

52

32

12

69.

< 53.4 inches

hs55d 5 116 logs55 1 40d 2 176

h 5 116 logsa 1 40d 2 176 70.

< 27.16 miles

s 5 25 213 lns10y12d

ln 371.

log4 9 5ln 9

ln 4< 1.585

log4 9 5log 9

log 4< 1.585


72.

log12 200 5ln 200

ln 12 < 2.132

log12 200 5log 200

log 12 < 2.132 73.

log1y2 5 5ln 5

lns1y2d< 22.322

log1y2 5 5log 5

logs1y2d< 22.322 74.

log3 0.28 5ln 0.28

ln 3< 21.159

log3 0.28 5log 0.28

log 3< 21.159

75.

< 1.255

5 log 2 1 2 log 3

log 18 5 logs2 ? 32d 76.

< 23.585

5 22 log2 22

2 log2 3 5 22 2log 3

log 2

log2 1

125 log2 1 2 log2 12 5 0 2 logs22

? 3d

77.

5 2 ln 2 1 ln 5 < 2.996

ln 20 5 lns22? 5d 78.

< 22.90

5 ln 3 2 4

ln 3e245 ln 3 1 ln e24 79.

5 1 1 2 log5 x

log5 5x25 log5 5 1 log5 x

2

80.

5 log 7 1 4 log x

log10 7x45 log 7 1 log x

4 81.

5 1 1 log3 2 21

3 log3 x

5 log3 3 1 log3 2 21

3 log3 x

5 log3s3 ? 2d 2 log3 x1y3

log3 6

3!x5 log3 6 2 log3

3!x 82.

51

2 log7 x 2 log7 4

5 log7 x1y2

2 log7 4

log7 !x

45 log7 !x 2 log7 4

83.

5 2 ln x 1 2 ln y 1 ln z

ln x2y2z 5 ln x21 ln y2

1 ln z 84.

5 ln 3 1 ln x 1 2 ln y

ln 3xy25 ln 3 1 ln x 1 ln y2

85.

5 lnsx 1 3d 2 ln x 2 ln y

5 lnsx 1 3d 2 fln x 1 ln yg

ln1x 1 3

xy 2 5 lnsx 1 3d 2 ln xy 86.

5 2 lnsy 2 1d 2 ln 16, y > 1

5 2 lnsy 2 1d 2 2 ln 4

ln1y 2 1

4 22

5 2 ln1y 2 1

4 2

87. log2 5 1 log2 x 5 log2 5x 88.

5 log6 y

z2

log6 y 2 2 log6 z 5 log6 y 2 log6 z2

89. ln x 21

4 ln y 5 ln x 2 ln 4!y 5 ln1 x

4!y2 90.

5 ln x3sx 1 1d2

3 ln x 1 2 lnsx 1 1d 5 ln x31 lnsx 1 1d2

91.

5 log8sy7 3!x 1 4d

1

3 log8sx 1 4d 1 7 log8 y 5 log8

3!x 1 4 1 log8 y7 92.

5 log 1

x2sx 1 6d5

5 log x22

sx 1 6d5

22 log x 2 5 logsx 1 6d 5 log x222 logsx 1 6d5


94.

5 ln sx 2 2d5

x3sx 1 2d

5 lnsx 2 2d52 ln x3sx 1 2d

5 lnsx 2 2d52 flnsx 1 2d 1 ln x3g

5 lnsx 2 2d 2 lnsx 1 2d 2 3 ln x 5 lnsx 2 2d52 lnsx 1 2d 2 ln x3

95.

(a) Domain:

(b)

Vertical asymptote: h 5 18,000

0 20,000

0

100

0 ≤ h < 18,000

t 5 50 log 18,000

18,000 2 h

(c) As the plane approaches its absolute ceiling, it climbs at

a slower rate, so the time required increases.

(d) minutes50 log 18,000

18,000 2 4000< 5.46

96. Using a calculator gives

s 5 84.66 1 s211 ln td.

97.

x 5 3

8x5 83

8x5 512 98.

x 5 23

6x5 623

6x5

1216 99.

x 5 ln 3

ex5 3

100.

x 5 ln 6 < 1.792

ln ex5 ln 6

ex5 6 101.

x 5 425 16

log4 x 5 2 102.

x 516

6log6 x 5 621

log6 x 5 21 103.

x 5 e4

ln x 5 4

104.

x 5 e23 < 0.0498

ln x 5 23 105.

x 5 ln 12 < 2.485

ln ex5 ln12

ex5 12 106.

x 5ln 25

3< 1.073

3x 5 ln 25

ln e3x5 ln 25

e3x5 25 107.

x 5 1 or x 5 3

0 5 sx 2 1dsx 2 3d

0 5 x22 4x 1 3

4x 5 x21 3

e4x5 ex2

13

108.

x 5sln 40d 2 2

3< 0.563

3x 1 2 5 ln 40

ln e3x125 ln 40

e3x125 40

14e3x125 560 109.

x < 4.459

5log 22

log 2 or

ln 22

ln 2

x 5 log2 22

2x5 22

2x1 13 5 35 110.

x 5ln 20

ln 6< 1.672

x 5 log6 20

log6 6x

5 log6 20

6x5 20

6x2 28 5 28

93.

5 ln !2x 2 1

sx 1 1d2

1

2 lns2x 2 1d 2 2 lnsx 1 1d 5 ln!2x 2 1 2 lnsx 1 1d2


113.

x 5 ln 2 < 0.693 x 5 ln 5 < 1.609

ln ex5 ln 2 ln e x

5 ln 5

ex5 2 or e x

5 5

sex2 2dsex

2 5d 5 0

e2x2 7ex

1 10 5 0 114.

x < 0.693 x < 1.386

x 5 ln 2 x 5 ln 4

ex5 2 or ex

5 4

sex2 2dsex

2 4d 5 0

e2x2 6ex

1 8 5 0

115.

Graph .

The x-intercepts are at

x < 0.392 and at x < 7.480.

y1 5 20.6x2 3x

−10

−10

10

1020.6x2 3x 5 0 116.

Graph .

The x-intercepts are at

x < 21.527.

x < 27.038 and at

y1 5 420.2x1 x

−3

−12 6

9420.2x1 x 5 0

117.

Graph and

The graphs intersect at

x < 2.447.

y2 5 12.

y1 5 25e20.3x

−6

16

−2

18

25e20.3x5 12 118.

Graph and

The graphs intersect at

x < 0.676.

y2 5 9.y1 5 4e1.2x

6−6

12

−2

4e1.2x5 9

119.

x 5e8.2

3< 1213.650

3x 5 e8.2

eln 3x5 e8.2

ln 3x 5 8.2 120.

x 5e7.2

5< 267.886

5x 5 e7.2

ln 5x 5 7.2 121.

x 51

4e7.5 < 452.011

4x 5 e7.5

eln 4x5 e7.5

ln 4x 515

2

2 ln 4x 5 15

122.

x 5e15y4

3< 14.174

3x 5 e15y4

ln 3x 515

4

4 ln 3x 5 15 123.

x 5 3e2 < 22.167

x

35 e2

elnsxy3d5 e2

ln x

35 2

ln x 2 ln 3 5 2 124.

x 5 e62 8 < 395.429

x 1 8 5 e6

lnsx 1 8d 5 6

1

2 lnsx 1 8d 5 3

ln!x 1 8 5 3

111.

x 5ln 17

ln 5< 1.760

x ln 5 5 ln 17

ln 5x5 ln 17

5x5 17

24s5xd 5 268 112.

x 5ln 95

ln 12< 1.833

x ln 12 5 ln 95

ln 12x5 ln 95

12x5 95

2s12xd 5 190


127.

Since is not in the domain of or of

it is an extraneous solution. The equation

has no solution.

log8sx 2 2d,log8sx 2 1dx 5 0

x 5 0

x25 0

x21 x 2 2 5 x 2 2

sx 2 1dsx 1 2d 5 x 2 2

x 2 1 5x 2 2

x 1 2

log8sx 2 1d 5 log81x 2 2

x 1 22log8sx 2 1d 5 log8sx 2 2d 2 log8sx 1 2d 128.

Quadratic Formula

Only is a valid solution.x 5 22 1 !6 < 0.449

x 5 22 ± !6,

0 5 x21 4x 2 2

x 1 2 5 x21 5x

x 1 2

x5 x 1 5

log61x 1 2

x 2 5 log6sx 1 5d

log6sx 1 2d 2 log6 x 5 log6sx 1 5d

129.

x 5 0.900

1 2110 5 x

1 2 x 5 1021

logs1 2 xd 5 21 130.

x 5 2104

2x 5 100 1 4

2x 2 4 5 102

logs2x 2 4d 5 2

131.

Graph

The graphs intersect at approximately

The solution of the equation is x < 1.643.

s1.643, 8d.

−9 9

−2

10

(1.64, 8)

y1 5 2 lnsx 1 3d 1 3x and y2 5 8.

2 lnsx 1 3d 1 3x 5 8 132.

Graph

The x-intercepts are at x 5 0, x < 0.416, and x < 13.627.

−4

−8 16

12

y1 5 6 logsx21 1d 2 x.

6 logsx21 1d 2 x 5 0

133.

Graph

The graphs do not intersect. The equation has no solution.

−6

11

−1

12

y1 5 4 lnsx 1 5d 2 x and y2 5 10.

4 lnsx 1 5d 2 x 5 10

125.

x 5 e42 1 < 53.598

x 1 1 5 e4

elnsx11d5 e4

lnsx 1 1d 5 4

1

2 lnsx 1 1d 5 2

ln!x 1 1 5 2 126.

x 5 5e4 < 272.991

x

55 e4

ln x

55 4

ln x 2 ln 5 5 4


134.

Let

The x-intercepts are at x < 23.990 and x < 1.477.

−4

−8 16

12

y1 5 x 2 2 logsx 1 4d.

x 2 2 logsx 1 4d 5 0 135.

t 5ln 3

0.0725< 15.2 years

ln 3 5 0.0725t

ln 3 5 ln e0.0725t

3 5 e0.0725t

3s7550d 5 7550e0.0725t

136.

d 5 10s218y93d < 220.8 miles

logsdd 5218

93

218 5 93 logsdd

283 5 93 logsdd 1 65

S 5 93 logsdd 1 65 137.

Exponential decay model

Matches graph (e).

y 5 3e22xy3

138.

Exponential growth model

Matches graph (b).

y 5 4e2xy3 139.

Logarithmic model

Vertical asymptote:

Graph includes

Matches graph (f).

s22, 0d

x 5 23

y 5 lnsx 1 3d

140.

Logarithmic model

Vertical asymptote:

Matches graph (d).

x 5 23

y 5 7 2 logsx 1 3d 141.

Gaussian model

Matches graph (a).

y 5 2e2sx14d2y3 142.

Logistics growth model

Matches graph (c).

y 56

1 1 2e22x

143.

Thus, y < 2e0.1014x.

ln 1.5 5 4b ⇒ b < 0.1014

1.5 5 e4b

3 5 2ebs4d

2 5 aebs0d ⇒ a 5 2

y 5 aebx 144.

y 51

2e0.4605x

b < 0.4605

ln 10

55 b

ln 10 5 5b

10 5 e5b

5 51

2ebs5d

1

25 aebs0d

⇒ a 51

2

y 5 aebx


145.

According to this model, the population of South

Carolina will reach 4.5 million during the year 2008.

t 5lns4500y3499d

0.0135< 18.6 years

ln14500

34992 5 0.0135t

4500

34995 e0.0135t

4500 5 3499e0.0135t

4.5 million 5 4500 thousand

P 5 3499e0.0135t 146.

When we have

After 5000 years, approximately 98.6% of the

radioactive uranium II will remain.

y 5 Ceflns1y2dy250,000gs5000d < 0.986C 5 98.6%C.

t 5 5000,

k 5lns1y2d250,000

ln 1

25 250,000k

ln 1

25 ln es250,000dk

1

2C 5 Ces250,000dk

y 5 Cekt

150.

(a) When

weeks t 5lns107y270d

20.12< 7.7

20.12t 5 ln 107

270

e20.12t5

107

270

5.4e20.12t5

107

50

1 1 5.4e20.12t5

157

50

50 5157

1 1 5.4e20.12t

N 5 50:

N 5157

1 1 5.4e20.12t

(b) When

weeks t 5lns82y405d

20.12< 13.3

20.12t 5 ln 82

405

e20.12t5

82

405

5.4e20.12t5

82

75

1 1 5.4e20.12t5

157

75

75 5157

1 1 5.4e20.12t

N 5 75:

147. (a)

(b)

< $11,486.98

A 5 10,000e0.138629

5 13.8629%

r < 0.138629

ln 2

55 r

ln 2 5 5r

2 5 e5r

20,000 5 10,000ers5d 148. and so

and:

The population one year ago:

5 1243 bats

Ns4d 5 2000e20.11889s4d

k 5lns7y10d

35 20.11889

3k 5 ln1 7

102

7

105 e3k

1400 5 2000e3k

N 5 2000ekt

N3 5 1400N0 5 2000 149.

(a) Graph

(b) The average test score is 71.

40 100

0

0.05

y1 5 0.0499e2sx271d2y128.

40 ≤ x ≤ 100

y 5 0.0499e2sx271d2y128,

Problem Solving for Chapter 3 329

151.

I 5 1023.5 wattycm2

1012.5 5I

10216

12.5 5 log1 I

102162

125 5 10 log1 I

102162

b 5 10 log1 I

102162 152.

(a)

(b)

(c)

I 5 109.1 < 1,258,925,412

log I 5 9.1

I 5 106.85 < 7,079,458

log I 5 6.85

I 5 108.4 < 251,188,643

log I 5 8.4

R 5 log I since I0 5 1.

153. True. By the inverse properties, logb b2x 5 2x. 154. False. ln x 1 ln y 5 lnsxyd Þ lnsx 1 yd

155. Since graphs (b) and (d) represent exponential decay, b and d are negative.

Since graph (a) and (c) represent exponential growth, a and c are positive.

Problem Solving for Chapter 3

1.

The curves and cross the line

From checking the graphs it appears that will cross

for 0 ≤ a ≤ 1.44.y 5 ax

y 5 x

y 5 x.y 5 1.2xy 5 0.5x

y4 5 x

y3 5 2.0x

y2 5 1.2x

y1 5 0.5x

y

x

−4 −3 −2 −1 1 2

7

6

5

4

3

2

3 4

y1

y3

y4

y2

y 5 ax 2.

The function that increases at the fastest rate for “large”

values of x is (Note: One of the intersection

points of and is approximately

and past this point This is not shown on the

graph above.)

ex> x3.

s4.536, 93dy 5 x3y 5 ex

y1 5 ex.

y5 5 |x| y4 5 !x

y3 5 x3

y2 5 x2

0

0 6

24

y1

y5

y3 y2

y4

y1 5 ex

3. The exponential function, increases at a faster rate

than the polynomial function y 5 xn.

y 5 ex, 4. It usually implies rapid growth.

5. (a)

(b)

5 f f sxdg2

5 saxd2

f s2xd 5 a2x

5 f sud ? f svd

5 au ? av

f su 1 vd 5 au1v 6.

5 1

54

4

5 1e2x 1 2 1 e22x

4 2 2 1e2x 2 2 1 e22x

4 2

f f sxdg2 2 fg sxdg2 5 1ex 1 e2x

2 22

2 1ex 2 e2x

2 22

7. (a)

−2

−6 6

6

y = ex y1

(b)y = ex

y2

−2

−6 6

6 (c)y = ex

y3

−2

−6 6

6


8.

As more terms are added, the polynomial approaches

ex 5 1 1x

1!1

x2

2!1

x3

3!1

x4

4!1

x5

5!1 . . .

ex.

−2

−6 6

6

y = exy4

y4 5 1 1x

1!1

x2

2!1

x3

3!1

x4

4!9.

Quadratic Formula

Choosing the positive quantity for we have

Thus, f 21sxd 5 ln1x 1!x2 1 4

2 2.y 5 ln1x 1 !x2 1 4

2 2.

ey

ey 5x ± !x2 1 4

2

e2y 2 xey 2 150

xey 5 e2y 2 1

x 5e2y 2 1

ey

x 5 ey 2 e2y

y 5 ex 2 e2x

f sxd 5 ex 2 e2x

10.

y 5 loga1x 1 1

x 2 12 5

ln1x 1 1

x 2 12ln a

5 f 21sxd

ay 5x 1 1

x 2 1

aysx 2 1d 5 x 1 1

xay 2 ay 5 x 1 1

xsay 2 1d 5 ay 1 1

x 5ay 1 1

ay 2 1

f sxd 5ax 1 1

ax 2 1, a > 0, a Þ 1 11. Answer (c).

The graph passes through and neither (a) nor (b) pass

through the origin. Also, the graph has y-axis symmetry and

a horizontal asymptote at y 5 6.

s0, 0d

y 5 6s1 2 e2x2y2d

12. (a) The steeper curve represents the investment earning compound interest,

because compound interest earns more than simple interest. With simple

interest there is no compounding so the growth is linear.

(b) Compound interest formula:

Simple interest formula:

(c) One should choose compound interest since the earnings would be higher.

A 5 Prt 1 P 5 500s0.07dt 1 500

A 5 500s1 10.07

1 ds1dt5 500s1.07dt

Time (in years)

Gro

wth

of

inves

tmen

t(i

n d

oll

ars)

A

t5 10 15 20 25

1000

2000

3000

4000

Compounded Interest

Simple Interest

30

13. and

t 5ln c1 2 ln c2

fs1yk2d 2 s1yk1dg lns1y2d

ln c1 2 ln c2 5 t1 1

k2

21

k12 ln11

22

ln1c1

c22 5 1 t

k2

2t

k12 ln11

22

c1

c2

5 11

22styk22tyk1d

c111

22tyk1

5 c211

22tyk2

y2 5 c211

22tyk2

y1 5 c111

22tyk1

14. through and

B 5 500afs1y2d logas2y5dgt 5 500 faloga s2y5dgty2 5 50012

52ty2

1

2 loga12

52 5 k

loga12

52 5 2k

2

55 a2k

200 5 500aks2d

B0 5 500

s2, 200ds0, 500dB 5 B0akt

y

x−4 −3 −2 −1 1 2

4

3

2

1

−4

3 4

Problem Solving for Chapter 3 331

15. (a)

(b)

(c)

(d) Both models appear to be “good fits” for the data, but

neither would be reliable to predict the population of

the United States in 2010. The exponential model

approaches infinity rapidly.

200,000

0 85

2,900,000

y2

y1

y < 400.88t 2 2 1464.6t 1 291,782

y < 252.606s1.0310dt16. Let and Then and

1 1 loga 1

b5

loga x

logayb x

1 1 loga 1

b5

m

n

loga 1

b5

m

n2 1

amyn21 51

b

amyn 5a

b

am 5 1a

b2n

x 5 saybdn.

x 5 amlogayb x 5 n.loga x 5 m

17.

x 5 1 or x 5 e2

ln x 5 0 or ln x 5 2

ln xsln x 2 2d 5 0

sln xd2 2 2 ln x 5 0

sln xd2 5 ln x2

18.

(a) (b) (c)y = ln xy3

−4

−3 9

4

y = ln x

y2

−4

−3 9

4

−4

−3 9

4

y = ln xy1

y3 5 sx 2 1d 212sx 2 1d2 1

13sx 2 1d3

y2 5 sx 2 1d 212sx 2 1d2

y1 5 x 2 1

y 5 ln x

19.

The pattern implies that

ln x 5 sx 2 1d 212sx 2 1d2 1

13sx 2 1d3 2

14sx 2 1d4 1 . . . .

−4

−3 9

4

y = ln x

y4

y4 5 sx 2 1d 212sx 2 1d2 1

13sx 2 1d3 2

14sx 2 1d4

20.

Slope:

y-intercept: s0, ln ad

m 5 ln b

ln y 5 sln bdx 1 ln a

ln y 5 ln a 1 x ln b

ln y 5 ln a 1 ln bx

ln y 5 lnsabxd

y 5 abx 21.

ys300d 5 80.4 2 11 ln 300 < 17.7 ft3/min

0

100 1500

30

y 5 80.4 2 11 ln x

Slope:

y-intercept: s0, ln ad

m 5 b

ln y 5 b ln x 1 ln a

ln y 5 ln a 1 b ln x

ln y 5 ln a 1 ln x b

ln y 5 lnsaxbd

y 5 axb


22. (a) cubic feet per minute

(b)

x < 382 cubic feet of air space per child.

x 5 e65.4y11

ln x 565.4

11

11 ln x 5 65.4

15 5 80.4 2 11 ln x

450

305 15 (c) Total air space required: cubic feet

Let floor space in square feet and feet.

If the ceiling height is 30 feet, the minimum number of

square feet of floor space required is 382 square feet.

x 5 382

11,460 5 xs30d

V 5 xh

h 5 30x 5

382s30d 5 11,460

23. (a)

(b) The data could best be modeled by a logarithmic

model.

(c) The shape of the curve looks much more logarithmic

than linear or exponential.

(d)

(e) The model is a good fit to the actual data.

0

0 9

9

y < 2.1518 1 2.7044 ln x

0 9

0

9 24. (a)

(b) The data could best be modeled by an exponential

model.

(c) The data scatter plot looks exponential.

(d)

(e) The model graph hits every point of the scatter plot.

0

0 9

36

y < 3.114s1.341dx

0 9

0

36

25. (a)

(b) The data could best be modeled by a linear model.

(c) The shape of the curve looks much more linear than

exponential or logarithmic.

(d)

(e) The model is a good fit to the actual data.

0

0 9

9

y < 20.7884x 1 8.2566

0 9

0

9 26. (a)

(b) The data could best be modeled by a logarithmic

model.

(c) The data scatter plot looks logarithmic.

(d)

(e) The model graph hits every point of the scatter plot.

0

0 9

10

y < 5.099 1 1.92 lnsxd

0 9

0

10

Practice Test for Chapter 3 333

Chapter 3 Practice Test

1. Solve for x: x3@55 8.

2. Solve for x: 3x215

181.

3. Graph fsxd 5 22x.

4. Graph gsxd 5 ex1 1.

5. If $5000 is invested at 9% interest, find the amount after three years if the interest is compounded

(a) monthly. (b) quarterly. (c) continuously.

6. Write the equation in logarithmic form: 7225

149.

7. Solve for x: x 2 4 5 log2 164.

8. Given evaluate logb 4!8y25.logb 2 5 0.3562 and logb 5 5 0.8271,

9. Write as a single logarithm.5 ln x 212 ln y 1 6 ln z

10. Using your calculator and the change of base formula, evaluate log9 28.

11. Use your calculator to solve for N: log10 N 5 0.6646

12. Graph y 5 log4 x.

13. Determine the domain of f sxd 5 log3sx22 9d.

14. Graph y 5 lnsx 2 2d.

15. True or false:ln x

ln y5 lnsx 2 yd

16. Solve for x: 5x5 41

17. Solve for x: x 2 x25 log5

125

18. Solve for x: log2 x 1 log2sx 2 3d 5 2

19. Solve for x:ex

1 e2x

35 4

20. Six thousand dollars is deposited into a fund at an annual interest rate of 13%. Find the time required

for the investment to double if the interest is compounded continuously.

Documents

C H A P T E R 3 Exponential and Logarithmic Functions · 266 Chapter 3 Exponential and Logarithmic Functions 12. Asymptote: x −3 − 2−1 1 2 3 5 4 3 2 −1 y y 5 0 f sx d5 s1