EXAMPLE 1 Rewrite logarithmic equations Logarithmic FormExponential Form 2323 = 8 a. = 2 log 83 4040 = 1b. 4 log 1 = 0 = c. 12 log 121 = d. 1/4 log –14

EXAMPLE 1 Rewrite logarithmic equations

Logarithmic Form Exponential Form

23 = 8a. =2

log 8 3

40 = 1b. 4

log 1 = 0

=c. 12

log 12 1

=d. 1/4

log –14

121 = 12

4=–11

4

GUIDED PRACTICE for Example 1

Rewrite the equation in exponential form.

Logarithmic Form Exponential Form

34 = 811. =3

log 81 4

71 = 72. 7

log 7 = 1

=3. 14

log 1 0

=4. 1/2

log –532

140 = 1

32=–51

2

EXAMPLE 5 Use inverse properties

Simplify the expression.

a. 10log4 b. 5

log 25x

SOLUTION

Express 25 as a power with base 5.

a. 10log4 = 4

b. 5

log 25x = (52) x

5log

=5

log 52x

2x=

Power of a power property

blog xb = x

blog bx = x

EXAMPLE 6 Find inverse functions

Find the inverse of the function.

SOLUTION

b.

a. y = 6 x b. y = ln (x + 3)

a.

6log

From the definition of logarithm, the inverse ofy = 6 x is y = x.

Write original function.y = ln (x + 3)Switch x and y.x = ln (y + 3)

Write in exponential form.

Solve for y.

=ex (y + 3)

=ex – 3 y

ANSWER The inverse of y = ln (x + 3) is y = ex – 3.

GUIDED PRACTICE for Examples 5 and 6


SOLUTION

10. 8 8log x

8 8log x = x blog bb = x

11. 7

log 7–3x

SOLUTION

7log 7–3x = –3x

alog ax = x



SOLUTION

12. 2

log 64x

Express 64 as a power with base 2.2

log 64x = (26) x

2log

=2

log 26x

6x=

Power of a power property

blog bx = x

13. eln20

SOLUTION

eln20 = elog 20e = 20 elog xe = x


Find the inverse of14. y = 4 x

SOLUTION

From the definition of logarithm, the inverse of

4logy = 6 y = x.is

y = ln (x – 5).Find the inverse of15.

y = ln (x – 5)

SOLUTION

Write original function.

Switch x and y.x = ln (y – 5)Write in exponential form.

Solve for y.

=ex (y – 5)=ex + 5 y

ANSWER The inverse of y = ln (x – 5) is y = ex + 5.

EXAMPLE 7 Graph logarithmic functions

Graph the function.

SOLUTION

a. y =3

log x

Plot several convenient points, such as (1, 0), (3, 1), and (9, 2). The y-axis is a vertical asymptote.

From left to right, draw a curve that starts just to the right of the y-axis and moves up through the plotted points, as shown below.

EXAMPLE 7 Graph logarithmic functions

Graph the function.

SOLUTION

b. y =1/2

log x

Plot several convenient points, such as (1, 0), (2, –1), (4, –2), and (8, –3). The y-axis is a vertical asymptote.

From left to right, draw a curve that starts just to the right of the y-axis and moves down through the plotted points, as shown below.

EXAMPLE 8 Translate a logarithmic graph

SOLUTION

STEP 1

Graph . State the domain and range.y =2

log (x + 3) + 1

STEP 2

Sketch the graph of the parent function y = x, which passes through (1, 0), (2, 1), and (4, 2).

2log

Translate the parent graph left 3 units and up 1 unit. The translated graph passes through (–2, 1), (–1, 2), and (1, 3). The graph’s asymptote is x = –3. The domain is x > –3, and the range is all real numbers.


Graph the function. State the domain and range.

SOLUTION

16. y =5

log x

If x = 1 y = 0,x = 5 y = 1,x = 10 y = 2

Plot several convenient points, such as (1, 0), (5, 1), and (10, 2). The y-axis is a vertical asymptote.


From left to right, draw a curve that starts just to the right of the y-axis and moves up through the plotted points.

The domain is x > 0, and the range is all real numbers.



SOLUTION

17. y =1/3

log (x – 3)

domain: x > 3,

range: all real numbers



SOLUTION

18. y =4

log (x + 1) – 2

domain: x > 21,

range: all real numbers

EXAMPLE 2 Evaluate logarithms

4loga. 64

b. 5

log 0.2

Evaluate the logarithm.

blogTo help you find the value of y, ask yourself what

power of b gives you y.

SOLUTION

4 to what power gives 64?a. 4

log43 64, so= 3.=64

5 to what power gives 0.2?b. =5–1 0.2, so –1.0.25

log =

EXAMPLE 2 Evaluate logarithms

Evaluate the logarithm.

blogTo help you find the value of y, ask yourself what

power of b gives you y.

SOLUTION

=–31

5 125, so1/5

log 125 =–3.c. to what power gives 125?15

d. 36 to what power gives 6? 361/2 6, so36

log 6= =12

.

d. 36

log 6

c. 1/5

log 125

EXAMPLE 3 Evaluate common and natural logarithms

Expression Keystrokes Display

a. log 8

b. ln 0.3

Check

8

.3

0.903089987

–1.203972804

100.903 8

0.3e –1.204

EXAMPLE 4 Evaluate a logarithmic model

Tornadoes

The wind speed s (in miles per hour) near the center of a tornado can be modeled by

where d is the distance (in miles) that the tornado travels. In 1925, a tornado traveled 220 miles through three states. Estimate the wind speed near the tornado’s center.

93 log d + 65s =

EXAMPLE 4 Evaluate a logarithmic model

SOLUTION

= 93 log 220 + 65

Write function.

93(2.342) + 65

= 282.806

Substitute 220 for d.

Use a calculator.

Simplify.

The wind speed near the tornado’s center was about 283 miles per hour.

ANSWER

93 log d + 65s =

GUIDED PRACTICE for Examples 2, 3 and 4

Evaluate the logarithm. Use a calculator if necessary.

2 to what power gives 32?

25 32, so= 5.=2

log 32

27 to what power gives 3?13

271/3 3, so27

log 3= = .

2log5. 32

SOLUTION

27log6. 3

SOLUTION


Evaluate the logarithm. Use a calculator if necessary.

Expression Keystrokes Display

7. log 12

8. ln 0.75

Check

12

.75

1.079

–0.288

101.079 12

0.75e –0.288


WHAT IF? Use the function in Example 4 to estimate the wind speed near a tornado’s center if its path is 150 miles long.

9.

= 93 log 150 + 65

Write function.

93(2.1760) + 65

= 267

Substitute 150 for d.

Use a calculator.

Simplify.

93 log d + 65s =

SOLUTION

The wind speed near the tornado’s center is about 267 miles per hour.

ANSWER

EXAMPLE 1 Use properties of logarithms

b. 4

log 21

a. 4

log 37 = 3 –

4log 7

4log

= –0.612

=4

log (3 7)

= 34

log + 74

log

= 2.196

0.792 1.404–

0.792 1.404+

34

logUse 0.792 and4

log 7 1.404 to evaluate thelogarithm.

Quotient property

Simplify.

Use the given values of 34

log7.

4log

and

Write 21 as 3 7.

Product property


log7.

4log

and

Simplify.

EXAMPLE 1 Use properties of logarithms

c. 4

log 49

34

logUse 0.792 and4

log 7 1.404 to evaluate thelogarithm.

72=4

log

= 2.808

2(1.404)

Write 49 as 72

Power property

Use the given value of 7.4

log

Simplify.

4log= 2 7


2. 6

log 40

= 5 –6

log 86

log

= –0.263

=6

log (8 5)

= 86

log + 56

log

= 2.059

1.161 0.898+

Quotient property

Simplify.

0.898 1.161– Use the given values of 56

log8.

6log

and

Write 40 as 8 5.

Product property


log8.

6log

and

Simplify.

1. 586

log

56

logUse 0.898 and 8 1.161 to evaluate thelogarithm.

6log


56

logUse 0.898 and 8 1.161 to evaluate thelogarithm.

6log

6log3. 64 82=

6log

= 2.322

= 2 86

log

2(1.161)

Write 64 as 82

Power property


log

Simplify.

4. 6

log 125 53=6

log

= 2.694

= 3 56

log

3(0.898)

Write 125 as 53

Power property


log

Simplify.

EXAMPLE 2 Expand a logarithmic expression

Power property

Expand6

log 5x3

y

6log 5x3

yQuotient property

= 56

log x3 y6

log –6

log+

= 56

log x y6

log –6

log+ 3

Product property

SOLUTION

= 5x3 y6

log –6

log

EXAMPLE 3 Standardized Test Practice

Quotient property

Product property

Simplify.

SOLUTION

–log 9 + 3log2 log 3 = –log 9 + log23 log 3 Power property

= 24log

= log (9 )23 – log 3

= log 9 23

3

The correct answer is D.ANSWER


Expand5. log 3x4 .

SOLUTION

= log 3 + log x4log 3x4

= log 3 + 4 log x Power property

Product property


SOLUTION

Quotient property

Product property

Simplify.

= –ln 4 + ln 33 ln 12 Power property

= ln (4 )33 – ln 12

= ln 4 33

12

= ln 9

Condense ln 4 + 3 ln 3 – ln 12.6.

ln 4 + 3 ln 3 – ln 12

EXAMPLE 4 Use the change-of-base formula

SOLUTION

3log 8Evaluate using common logarithms and natural

logarithms.

Using common logarithms:

Using natural logarithms:

3log 8 =

log 8log 3

0.90310.4771

1.893

3log 8 =

ln 8ln 3

2.07941.0986

1.893

EXAMPLE 5 Use properties of logarithms in real life

For a sound with intensity I (in watts per square meter), the loudness L(I) of the sound (in decibels) is given by the function

= logL(I) 10 I

0I

Sound Intensity

0Iwhere is the intensity of a barely audible sound

(about watts per square meter). An artist in a recording studio turns up the volume of a track so that the sound’s intensity doubles. By how many decibels does the loudness increase?

10–12

EXAMPLE 5 Use properties of logarithms in real life

Product property

Simplify.

SOLUTION

Let I be the original intensity, so that 2I is the doubled intensity.

Increase in loudness = L(2I) – L(I)

= log10 I

0I

log10 2I

0I

–

I

0I

2I

0I= 10 loglog –

= 210 log log I

0I

–log I

0I

+

ANSWER The loudness increases by about 3 decibels.

10 log 2=

3.01

Write an expression.

Substitute.

Distributive property

Use a calculator.


Use the change-of-base formula to evaluate the logarithm.

5log 87.

SOLUTION

5log 8 =

log 8log 5

0.90310.6989

1.292

8log 148.

SOLUTION

8log 14 =

log 14log 8

1.1460.9031

1.269


Use the change-of-base formula to evaluate the logarithm.

26log 99.

SOLUTION

SOLUTION

=log 30log 12

1.47771.076

1.369

0.95421.4149

0.67426

log 9

10. 12

log 30

12log 30

=log 9

log 26


WHAT IF? In Example 5, suppose the artist turns up the volume so that the sound’s intensity triples. By how many decibels does the loudness increase?

11.

SOLUTION

Let I be the original intensity, so that 3I is the tripled intensity.

= logL(I) 10 I

0I


Product property

Simplify.

=Increase in loudness L(3I) – L(I)

= log10 I

0I

log10 3I

0I

–

= 310 log log I

0I

–log I

0I

+

ANSWER

The loudness increases by about 4.771 decibels.

10 log 3=

4.771

Write an expression.

Substitute.

Distributive property

Use a calculator.

= 10 log I

0I

log 3I

0I

–

Documents

EXAMPLE 1 Rewrite logarithmic equations Logarithmic FormExponential Form 2323 = 8 a. = 2 log 83 4040 = 1b. 4 log 1 = 0 = c. 12 log 121 = d. 1/4 log –14