5 Exponential Functions Logarithmic Functions Compound Interest Differentiation of Exponential Functions Exponential Functions as Mathematical

55

Exponential FunctionsExponential Functions Logarithmic FunctionsLogarithmic Functions Compound InterestCompound Interest Differentiation of Exponential FunctionsDifferentiation of Exponential Functions Exponential Functions as Mathematical ModelsExponential Functions as Mathematical Models

Exponential and Logarithmic FunctionsExponential and Logarithmic Functions

5.15.1Exponential FunctionsExponential Functions

x

y

– 2 2

4

2

f(x) = (1/2)x

f(x) = 2x

Exponential FunctionExponential Function

The function defined byThe function defined by

is called an is called an exponential functionexponential function with with basebase bb and and exponentexponent xx..

The The domaindomain of of ff is the set of is the set of all real numbersall real numbers..

( ) ( 0, 1) xf x b b b ( ) ( 0, 1) xf x b b b

ExampleExample

The The exponential functionexponential function with with basebase 2 2 is the functionis the function

with with domaindomain (–(– , , )). . Find the Find the valuesvalues of of ff((xx)) for selected values of for selected values of x x follow: follow:

( ) 2xf x ( ) 2xf x

(3)f (3)f

3

2f

3

2f

(0)f (0)f

32 832 8

3/2 1/22 2 2 2 2 3/2 1/22 2 2 2 2

02 102 1

ExampleExample

The The exponential functionexponential function with with basebase 2 2 is the functionis the function

with with domaindomain (–(– , , )). . Find the Find the valuesvalues of of ff((xx)) for selected values of for selected values of x x follow: follow:

( ) 2xf x ( ) 2xf x

( 1)f ( 1)f

2

3f

2

3f

1 12

2 1 1

22

2/32/3 3

1 12

2 4 2/3

2/3 3

1 12

2 4

Laws of ExponentsLaws of Exponents

Let Let aa and and bb be be positive numberspositive numbers and let and let xx andand y y be be real numbersreal numbers. Then,. Then,

1.1.

2.2.

3.3.

4.4.

5.5.

x y x yb b b x y x yb b b x

x yy

bb

b

xx y

y

bb

b

yx xyb b yx xyb b

x x xab a b x x xab a bx x

x

a a

b b

x x

x

a a

b b

ExamplesExamples

Let Let ff((xx) = 2) = 222xx – 1 – 1. . Find the valueFind the value of of xx for which for which ff((xx) = 16) = 16..

SolutionSolution We want to solve the equationWe want to solve the equation

2222xx – 1 – 1 = 16 = 2= 16 = 244

But this equation holds if and only if But this equation holds if and only if

22xx – 1 = 4 – 1 = 4

giving giving xx = = ..

Example 2, page 331

5

2

ExamplesExamples

Sketch the graphSketch the graph of the exponential function of the exponential function ff((xx) = 2) = 2xx. .

SolutionSolution First, recall that the First, recall that the domaindomain of this function is the of this function is the set of set of

real numbersreal numbers.. Next, putting Next, putting xx = 0 = 0 gives gives yy = 2 = 200 = 1 = 1, which is the , which is the yy-intercept-intercept..

(There is no (There is no xx-intercept-intercept, since there is no value of , since there is no value of xx for for which which yy = 0 = 0))

Example 3, page 331

ExamplesExamples


SolutionSolution Now, consider a few values for Now, consider a few values for xx::

Note that Note that 22xx approaches zeroapproaches zero as as xx decreases without bounddecreases without bound::✦ There is a There is a horizontal asymptotehorizontal asymptote at at yy = 0 = 0..

Furthermore, Furthermore, 22xx increases without boundincreases without bound when when xx increases increases without boundwithout bound..

Thus, the Thus, the range range of of ff is the is the intervalinterval (0, (0, ))..

xx – – 5 5 – – 44 – – 33 – – 22 – – 11 00 11 22 33 44 55

yy 1/321/32 1/161/16 1/81/8 1/41/4 1/21/2 11 22 44 88 1616 3232

Example 3, page 331

ExamplesExamples


SolutionSolution Finally, Finally, sketchsketch the graph: the graph:

xx

yy

– – 2 2 22

44

22

ff((xx) = 2) = 2xx

Example 3, page 331

ExamplesExamples

Sketch the graphSketch the graph of the exponential function of the exponential function ff((xx) = (1/2)) = (1/2)xx. .

SolutionSolution First, recall again that the First, recall again that the domaindomain of this function is the of this function is the

set of real numbersset of real numbers.. Next, putting Next, putting xx = 0 = 0 gives gives yy = (1/2) = (1/2)00 = 1 = 1, which is the , which is the

yy-intercept-intercept..

(There is no (There is no xx-intercept-intercept, since there is no value of , since there is no value of xx for for which which yy = 0 = 0))

Example 4, page 332

ExamplesExamples


SolutionSolution Now, consider a few values for Now, consider a few values for xx::

Note that Note that (1/2)(1/2)xx increasesincreases without bound without bound when when xx decreasesdecreases without boundwithout bound..

Furthermore, Furthermore, (1/2)(1/2)xx approaches zeroapproaches zero as as xx increasesincreases without without boundbound: there is a : there is a horizontal asymptotehorizontal asymptote at at yy = 0 = 0..

As before, the As before, the range range of of ff is the is the intervalinterval (0, (0, ))..

xx – – 5 5 – – 44 – – 33 – – 22 – – 11 00 11 22 33 44 55

yy 3232 1616 88 44 22 11 1/21/2 1/41/4 1/81/8 1/161/16 1/321/32

Example 4, page 332

ExamplesExamples


SolutionSolution Finally, Finally, sketchsketch the graph: the graph:

xx

yy

– – 2 2 22

44

22

ff((xx) = (1/2)) = (1/2)xx

Example 4, page 332

ExamplesExamples


SolutionSolution Note the Note the symmetrysymmetry between the two functions: between the two functions:

x

y

– 2 2

4

2

ff((xx) = (1/2)) = (1/2)xx

ff((xx) = 2) = 2xx

Example 4, page 332

Properties of Exponential FunctionsProperties of Exponential Functions

The The exponential functionexponential function yy = = bbxx ( (bb > 0, > 0, bb ≠ 1)≠ 1) has has the following properties:the following properties:

1.1. Its Its domaindomain is is (–(– , , ))..

2.2. Its Its rangerange is is (0, (0, ))..

3.3. Its graph Its graph passes throughpasses through the point the point (0, 1)(0, 1)

4.4. It is It is continuouscontinuous on on (–(– , , ))..

5.5. It is It is increasingincreasing on on (–(– , , )) if if bb > 1 > 1 and and decreasingdecreasing on on (–(– , , )) if if bb < 1 < 1..

The Base The Base ee

Exponential functionsExponential functions to the to the basebase ee, where , where ee is an is an irrational numberirrational number whose value is whose value is 2.7182818… 2.7182818…, play an , play an important role in both theoretical and applied problems.important role in both theoretical and applied problems.

It can be shown thatIt can be shown that

1lim 1

m

me

m

1lim 1

m

me

m

ExamplesExamples

Sketch the graphSketch the graph of the exponential function of the exponential function ff((xx) = ) = eexx. .

SolutionSolution Since Since eexx > 0> 0 it follows that the graph of it follows that the graph of yy = = eexx is is similarsimilar to the to the

graph ofgraph of yy = 2 = 2xx.. Consider a few values for Consider a few values for xx::

xx – – 33 – – 22 – – 11 00 11 22 33

yy 0.050.05 0.140.14 0.370.37 11 2.722.72 7.397.39 20.0920.09

Example 5, page 333

55

33

11

ExamplesExamples

Sketch the graphSketch the graph of the exponential function of the exponential function ff((xx) = ) = eexx. .

SolutionSolution SketchingSketching the graph: the graph:

xx

yy

– – 33 – – 11 1 1 33

ff((xx) = ) = eexx

Example 5, page 333

ExamplesExamples

Sketch the graphSketch the graph of the exponential function of the exponential function ff((xx) = ) = ee–x–x. .

SolutionSolution Since Since ee–x–x > 0> 0 it follows that it follows that 0 < 1/0 < 1/ee < 1 < 1 and so and so

ff((xx) = ) = ee–x–x = 1/= 1/eexx = (1/= (1/ee))xx is an exponential function with is an exponential function with base base less thanless than 11..

Therefore, it has a graph Therefore, it has a graph similarsimilar to that of to that of yy = (1/2) = (1/2)xx.. Consider a few values for Consider a few values for xx::

xx – – 33 – – 22 – – 11 00 11 22 33

yy 20.0920.09 7.397.39 2.722.72 11 0.370.37 0.140.14 0.050.05

Example 6, page 333

5

3

1

ExamplesExamples

Sketch the graphSketch the graph of the exponential function of the exponential function ff((xx) = ) = ee–x–x. .

SolutionSolution SketchingSketching the graph: the graph:

x

y

– 3 – 1 1 3

ff((xx) = ) = ee–x–x

Example 6, page 333

5.25.2Logarithmic FunctionsLogarithmic Functions

1

x

y

1

y = ex

y = ln x

y = x

LogarithmsLogarithms

We’ve discussed We’ve discussed exponential equationsexponential equations of the form of the form

yy = = bbxx ( (bb > 0, > 0, bb ≠ 1)≠ 1) But what about But what about solvingsolving the same equation the same equation forfor yy?? You may recall that You may recall that yy is called the is called the logarithmlogarithm of of xx to the to the

base base bb, and is denoted , and is denoted loglogbbxx..

✦ Logarithm of Logarithm of x x to the base to the base bb

yy = log = logbbxx if and only ifif and only if xx = = bbyy ((xx > 0) > 0)

ExamplesExamples

Solve Solve loglog33xx = 4= 4 for for xx::

SolutionSolution By definition, By definition, loglog33xx = 4= 4 implies implies xx = 3 = 344 = 81 = 81. .

Example 2, page 338

ExamplesExamples

Solve Solve loglog161644 = = xx for for xx::

SolutionSolution loglog161644 = = xx is equivalent to is equivalent to 4 = 164 = 16xx = (4 = (422))xx = 4= 422xx, or , or 4411 = 4 = 422xx,,

from which we deduce thatfrom which we deduce that

2 1

1

2

x

x

2 1

1

2

x

x

Example 2, page 338

ExamplesExamples

Solve Solve loglogxx88 = 3= 3 for for xx::

SolutionSolution By definition, we see that By definition, we see that loglogxx88 = 3 = 3 is equivalent to is equivalent to

3 38 2

2

x

x

3 38 2

2

x

x

Example 2, page 338

Logarithmic NotationLogarithmic Notation

log log xx = log= log1010 xx Common logarithmCommon logarithm

ln ln xx = log= logee xx Natural logarithmNatural logarithm

Laws of LogarithmsLaws of Logarithms

If If mm and and nn are are positive numberspositive numbers, then, then

1.1.

2.2.

3.3.

4.4.

5.5.

log log logb b bmn m n log log logb b bmn m n

log log logb b b

mm n

n log log logb b b

mm n

n

log lognb bm n mlog lognb bm n m

log 1 0b log 1 0b

log 1b b log 1b b

ExamplesExamples

Given that Given that log 2 log 2 ≈ 0.3010≈ 0.3010, , log 3 ≈ 0.4771log 3 ≈ 0.4771, and , and log 5 ≈ 0.6990log 5 ≈ 0.6990, , use the use the laws of logarithmslaws of logarithms to find to find

log15log15 log3 5

log3 log5

0.4771 0.6990

1.1761

log3 5

log3 log5

0.4771 0.6990

1.1761

Example 4, page 339

ExamplesExamples


log7.5log7.5 log(15 / 2)

log(3 5 / 2)

log3 log5 log 2

0.4771 0.6990 0.3010

0.8751

log(15 / 2)

log(3 5 / 2)

log3 log5 log 2

0.4771 0.6990 0.3010

0.8751

Example 4, page 339

ExamplesExamples


log81log81 4log3

4log3

4(0.4771)

1.9084

4log3

4log3

4(0.4771)

1.9084

Example 4, page 339

ExamplesExamples


log50log50 log5 10

log5 log10

0.6990 1

1.6990

log5 10

log5 log10

0.6990 1

1.6990

Example 4, page 339

ExamplesExamples

ExpandExpand and and simplifysimplify the expression: the expression:

2 33log x y2 33log x y 2 3

3 3

3 3

log log

2log 3log

x y

x y

2 33 3

3 3

log log

2log 3log

x y

x y

Example 5, page 340

ExamplesExamples


2

2

1log

2x

x 2

2

1log

2x

x

22 2

22 2

22

log 1 log 2

log 1 log 2

log 1

xx

x x

x x

22 2

22 2

22

log 1 log 2

log 1 log 2

log 1

xx

x x

x x

Example 5, page 340

ExamplesExamples


2 2 1ln

x

x x

e

2 2 1ln

x

x x

e

2 2 1/2

2 2 1/2

2

2

( 1)ln

ln ln( 1) ln

12ln ln( 1) ln

21

2ln ln( 1)2

x

x

x x

e

x x e

x x x e

x x x

2 2 1/2

2 2 1/2

2

2

( 1)ln

ln ln( 1) ln

12ln ln( 1) ln

21

2ln ln( 1)2

x

x

x x

e

x x e

x x x e

x x x

Example 5, page 340

Logarithmic FunctionLogarithmic Function

The function defined byThe function defined by

is called the is called the logarithmic functionlogarithmic function with with basebase bb.. The The domaindomain of of ff is the set of is the set of all positive numbersall positive numbers..

( ) log ( 0), 1)bf x x b b ( ) log ( 0), 1)bf x x b b

Properties of Logarithmic FunctionsProperties of Logarithmic Functions

The logarithmic function The logarithmic function

yy = log = logbbxx ((b b > 0, > 0, bb ≠ 1)≠ 1)

has the following has the following propertiesproperties::

1.1. Its Its domaindomain is is (0, (0, ))..

2.2. Its Its rangerange is is (–(– , , ))..

3.3. Its graph passes through the point Its graph passes through the point (1, 0)(1, 0)..

4.4. It is It is continuouscontinuous on on (0, (0, ))..

5.5. It is It is increasingincreasing on on (0, (0, )) if if b > 1b > 1 and and decreasingdecreasing on on (0, (0, )) if if b < 1b < 1..

ExampleExample

SketchSketch the graph of the function the graph of the function yy = ln = ln xx..SolutionSolution We first sketch the graph of We first sketch the graph of yy = = eexx..

11

xx

yy

11

yy = = eexx

yy = ln = ln xx

yy = = xx

The required graph is The required graph is the the mirror imagemirror image of the of the graph of graph of yy = = eexx with with respect to the line respect to the line y y == x x::

Example 6, page 341

Properties Relating Properties Relating Exponential and Logarithmic FunctionsExponential and Logarithmic Functions

Properties relating Properties relating eexx and and lnln x x::

eeln ln xx = = xx ((xx > 0) > 0)

ln ln eexx = = xx (for any real number (for any real number xx))

ExamplesExamples

Solve the equation Solve the equation 22eexx + 2 + 2 = 5 = 5..

SolutionSolution Divide both sidesDivide both sides of the equation by of the equation by 22 to obtain: to obtain:

Take the Take the natural logarithmnatural logarithm of of each sideeach side of the equation of the equation and and solvesolve::

2 52.5

2xe 2 5

2.52

xe

2ln ln 2.5

( 2) ln ln 2.5

2 ln 2.5

2 ln 2.5

1.08

xe

x e

x

x

x

2ln ln 2.5

( 2) ln ln 2.5

2 ln 2.5

2 ln 2.5

1.08

xe

x e

x

x

x

Example 7, page 342

ExamplesExamples

Solve the equation Solve the equation 5 ln 5 ln x x + 3 = 0+ 3 = 0..

SolutionSolution Add Add –– 33 to both sides to both sides of the equation and then of the equation and then divide both divide both

sidessides of the equation by of the equation by 55 to obtain: to obtain:

and so:and so:

5ln 3

3ln 0.6

5

x

x

5ln 3

3ln 0.6

5

x

x

ln 0.6

0.6

0.55

xe e

x e

x

ln 0.6

0.6

0.55

xe e

x e

x

Example 8, page 343

5.35.3Compound InterestCompound Interest

12

2 12 3

3 2

(1 )(1 ) [ (1 )](1 ) (1 )(1 ) [ (1 ) ](1 ) (1 )

A P iA A i P i i P iA A i P i i P i

First Period:

Second Period:

Third Period:

1

1(1 ) [ (1 ) ](1 ) (1 )n nn nA A i P i i P i

th Period:

n

12

2 12 3

3 2

(1 )(1 ) [ (1 )](1 ) (1 )(1 ) [ (1 ) ](1 ) (1 )


First Period:

Second Period:

Third Period:

1

1(1 ) [ (1 ) ](1 ) (1 )n nn nA A i P i i P i

th Period:

n

Compound InterestCompound Interest

Compound interest is a natural application of the Compound interest is a natural application of the exponential functionexponential function to to businessbusiness..

Recall that Recall that simple interestsimple interest is interest that is computed only is interest that is computed only on the on the original principaloriginal principal..

Thus, if Thus, if II denotes the denotes the interestinterest on a on a principalprincipal PP (in dollars) (in dollars) at an at an interest rateinterest rate of of rr per year for per year for tt years, then we have years, then we have

I I == Prt Prt The The accumulatedaccumulated amount amount AA, the sum of the , the sum of the principalprincipal and and

interestinterest after after tt years, is given by years, is given by

Simple interest formula(1 )

A P I P Prt

P rt

Simple interest formula(1 )

A P I P Prt

P rt


Frequently, interest earned is Frequently, interest earned is periodicallyperiodically added to the added to the principal and thereafter principal and thereafter earns interest itselfearns interest itself at the same at the same rate. This is called rate. This is called compound interestcompound interest..

Suppose Suppose $1000$1000 (the principal) is deposited in a bank for a (the principal) is deposited in a bank for a term of term of 33 years, earning interest at the rate of years, earning interest at the rate of 8%8% per year per year compounded annually.compounded annually.

Using the Using the simple interest formulasimple interest formula we see that the we see that the accumulated amount after the first year isaccumulated amount after the first year is

or or $1080$1080..

1 (1 )

1000[1 0.08(1)]

1000(1.08) 1080

A P rt

1 (1 )

1000[1 0.08(1)]

1000(1.08) 1080

A P rt


To find the accumulated amount To find the accumulated amount AA22 at the end of the at the end of the

second year, we use the second year, we use the simple interest formulasimple interest formula againagain, this , this time with time with PP = = AA11, obtaining:, obtaining:

or approximately or approximately $1166.40$1166.40..

2 1

2 2

(1 ) (1 )

1000[1 0.08(1)][1 0.08(1)]

1000(1 0.08) 1000(1.08) 1166.40

A P rt A rt

2 1

2 2

(1 ) (1 )

1000[1 0.08(1)][1 0.08(1)]

1000(1 0.08) 1000(1.08) 1166.40

A P rt A rt


We can use the We can use the simple interest formulasimple interest formula yet againyet again to find to find the accumulated amount the accumulated amount AA33 at the end of the third year: at the end of the third year:

or approximately or approximately $1259.71$1259.71..

3 2

2

3 3

(1 ) (1 )

1000[1 0.08(1)] [1 0.08(1)]

1000(1 0.08) 1000(1.08) 1259.71

A P rt A rt

3 2

2

3 3

(1 ) (1 )

1000[1 0.08(1)] [1 0.08(1)]

1000(1 0.08) 1000(1.08) 1259.71

A P rt A rt


Note that the accumulated amounts at the end of each year Note that the accumulated amounts at the end of each year have the following form:have the following form:

These observations suggest the following These observations suggest the following general rulegeneral rule::✦ If If PP dollars are dollars are investedinvested over a term of over a term of tt yearsyears earning earning

interestinterest at the rate of at the rate of rr per year per year compounded annuallycompounded annually, , then the then the accumulated amountaccumulated amount is is

1

22

33

1000(1.08)

1000(1.08)

1000(1.08)

A

A

A

1

22

33

1000(1.08)

1000(1.08)

1000(1.08)

A

A

A

1

22

33

(1 )

(1 )

(1 )

A P r

A P r

A P r

1

22

33

(1 )

(1 )

(1 )

A P r

A P r

A P r

or:

(1 )tA P r (1 )tA P r

Compounding More Than Once a YearCompounding More Than Once a Year

The formula The formula

was derived under the was derived under the assumptionassumption that interest was that interest was compoundedcompounded annuallyannually..

In practice, however, interest is usually In practice, however, interest is usually compoundedcompounded more more than once a yearthan once a year..

The interval of time between successive interest The interval of time between successive interest calculations is called the calculations is called the conversion periodconversion period..

(1 )tA P r (1 )tA P r


If interest at a nominal a rate of If interest at a nominal a rate of rr per year is per year is compoundedcompounded mm times a year on a times a year on a principalprincipal of of PP dollars, then the dollars, then the simple simple interest rate per conversion periodinterest rate per conversion period is is

For example, the For example, the nominal interestnominal interest rate is rate is 8%8% per year, and per year, and interest is interest is compounded quarterlycompounded quarterly, then, then

or or 2%2% per period. per period.

Annual interest rate

Periods per year

ri

m

Annual interest rate

Periods per year

ri

m

0.080.02

4

ri

m

0.080.02

4

ri

m


To find a general formula for the accumulated amount, we To find a general formula for the accumulated amount, we applyapply

repeatedlyrepeatedly with the interest rate with the interest rate i i == r/m r/m.. We see that the We see that the accumulated amountaccumulated amount at the at the end of each end of each

periodperiod is as follows: is as follows:

(1 )tA P r (1 )tA P r

12

2 12 3

3 2

(1 )(1 ) [ (1 )](1 ) (1 )(1 ) [ (1 ) ](1 ) (1 )


First Period:

Second Period:

Third Period:

1

1(1 ) [ (1 ) ](1 ) (1 )n nn nA A i P i i P i

th Period:

n

12

2 12 3

3 2

(1 )(1 ) [ (1 )](1 ) (1 )(1 ) [ (1 ) ](1 ) (1 )


First Period:

Second Period:

Third Period:

1

1(1 ) [ (1 ) ](1 ) (1 )n nn nA A i P i i P i

th Period:

n

Compound Interest FormulaCompound Interest Formula

wherewhere

AA == Accumulated amount at the end of Accumulated amount at the end of tt years years

PP == Principal Principal

rr == Nominal interest rate per year Nominal interest rate per year

mm == Number of conversion periods per year Number of conversion periods per year

tt == Term (number of years) Term (number of years)

1mt

rA P

m 1

mtr

A Pm

There are There are n n == mt mt periods in periods in tt years, so the years, so the accumulated accumulated amountamount at at the end ofthe end of tt yearyear is given by is given by

ExampleExample

Find the Find the accumulated amountaccumulated amount after after 3 3 years if years if $1000$1000 is is invested at invested at 8%8% per year per year compoundedcompoundeda.a. AnnuallyAnnuallyb.b. SemiannuallySemiannuallyc.c. QuarterlyQuarterlyd.d. MonthlyMonthlye.e. DailyDaily

Example 1, page 347

ExampleExample

SolutionSolution

a.a. Annually. Annually.

Here, Here, PP = 1000 = 1000, , rr = 0.08 = 0.08, , mm = 1 = 1, and , and tt = 3 = 3, so, so

or or $1259.71$1259.71..

(1)(3)

3

1

0.081000 1

1

1000(1.08)

1259.71

mtr

A Pm

(1)(3)

3

1

0.081000 1

1

1000(1.08)

1259.71

mtr

A Pm

Example 1, page 347

ExampleExample

SolutionSolution

b.b. Semiannually. Semiannually.


or or $1265.32$1265.32..

(2)(3)

6

1

0.081000 1

2

1000(1.04)

1265.32

mtr

A Pm

(2)(3)

6

1

0.081000 1

2

1000(1.04)

1265.32

mtr

A Pm

Example 1, page 347

ExampleExample

SolutionSolution

c.c. Quarterly. Quarterly.


or or $1268.24$1268.24..

(4)(3)

12

1

0.081000 1

4

1000(1.02)

1268.24

mtr

A Pm

(4)(3)

12

1

0.081000 1

4

1000(1.02)

1268.24

mtr

A Pm

Example 1, page 347

ExampleExample

SolutionSolution

d.d. Monthly. Monthly.


or or $1270.24$1270.24..

(12)(3)

36

1

0.081000 1

12

1000(1.00667)

1270.24

mtr

A Pm

(12)(3)

36

1

0.081000 1

12

1000(1.00667)

1270.24

mtr

A Pm

Example 1, page 347

ExampleExample

SolutionSolution

e.e. Daily. Daily.


or or $1271.22$1271.22..

(365)(3)

1095

1

0.081000 1

365

1000(1.00022)

1271.22

mtr

A Pm

(365)(3)

1095

1

0.081000 1

365

1000(1.00022)

1271.22

mtr

A Pm

Example 1, page 347

Effective Rate of InterestEffective Rate of Interest

The last example demonstrates that the The last example demonstrates that the interest actually interest actually earnedearned on an investment on an investment depends on the frequencydepends on the frequency with with which the interest is which the interest is compoundedcompounded..

For clarity when comparing interest rates, we can use For clarity when comparing interest rates, we can use what is called the what is called the effective rateeffective rate (also called the (also called the true ratetrue rate): ): ✦ This is the This is the simple interest ratesimple interest rate that would produce the that would produce the

same accumulated amountsame accumulated amount in in 11 year as the year as the nominal rate nominal rate compoundedcompounded mm timestimes a year. a year.

We want to We want to derive a relationderive a relation between the between the nominal nominal compounded ratecompounded rate and the and the effective rateeffective rate. .

Effective Rate of InterestEffective Rate of Interest

The accumulated amount after The accumulated amount after 11 year at a year at a simple interestsimple interest raterate rreffeff per year is per year is

The accumulated amount after The accumulated amount after 11 year at a year at a nominal interest nominal interest raterate rr per year per year compounded compounded mm times times a year is a year is

Equating the two expressions givesEquating the two expressions gives

eff(1 )A P r eff(1 )A P r

1m

rA P

m 1

mr

A Pm

Since t = 1

eff

eff

(1 ) 1

1 1

m

m

rP r P

m

rr

m

eff

eff

(1 ) 1

1 1

m

m

rP r P

m

rr

m

Effective Rate of Interest FormulaEffective Rate of Interest Formula

Solving the last equation for Solving the last equation for rreffeff we obtain the formula for we obtain the formula for

computing the effective rate of interest:computing the effective rate of interest:

eff 1 1m

rr

m

eff 1 1m

rr

m

wherewhere

rreffeff == Effective rate of interestEffective rate of interest

rr == Nominal interest rate per yearNominal interest rate per year

mm == Number of conversion periods per yearNumber of conversion periods per year

ExampleExample

Find the Find the effectiveeffective rate of interest rate of interest corresponding to a corresponding to a nominal ratenominal rate of of 8%8% per year per year compoundedcompoundeda.a. AnnuallyAnnuallyb.b. SemiannuallySemiannuallyc.c. QuarterlyQuarterlyd.d. MonthlyMonthlye.e. DailyDaily

Example 2, page 350

ExampleExample

SolutionSolution

a.a. Annually. Annually.

Let Let rr = 0.08 = 0.08 and and mm = 1 = 1. Then. Then

or or 8%8%..

1

eff

0.081 1

1

1.08 1

0.08

r

1

eff

0.081 1

1

1.08 1

0.08

r

Example 2, page 350

ExampleExample

SolutionSolution

b.b. Semiannually. Semiannually.


or or 8.16%8.16%..

2

eff

0.081 1

2

1.0816 1

0.0816

r

2

eff

0.081 1

2

1.0816 1

0.0816

r

Example 2, page 350

ExampleExample

SolutionSolution

c.c. Quarterly. Quarterly.


or or 8.243%8.243%..

4

eff

0.081 1

4

1.08243 1

0.08243

r

4

eff

0.081 1

4

1.08243 1

0.08243

r

Example 2, page 350

ExampleExample

SolutionSolution

d.d. Monthly. Monthly.


or or 8.3%8.3%..

12

eff

0.081 1

12

1.083 1

0.083

r

12

eff

0.081 1

12

1.083 1

0.083

r

Example 2, page 350

ExampleExample

SolutionSolution

e.e. Daily. Daily.


or or 8.328%8.328%..

365

eff

0.081 1

365

1.08328 1

0.08328

r

365

eff

0.081 1

365

1.08328 1

0.08328

r

Example 2, page 350

Effective Rate Over Several YearsEffective Rate Over Several Years

If the effective rate of interest If the effective rate of interest rreffeff is known, is known,

then the accumulated amount after then the accumulated amount after tt years on years on an investment of an investment of PP dollars may be more dollars may be more readily computed by using the formulareadily computed by using the formula

eff(1 )tA P r eff(1 )tA P r

Present ValuePresent Value

Consider the compound interest formula:Consider the compound interest formula:

The The principalprincipal PP is often referred to as the is often referred to as the present valuepresent value, , and the and the accumulated valueaccumulated value AA is called the is called the future valuefuture value, , since it is realized at a future date.since it is realized at a future date.

On occasion, an investor may wish to determine how much On occasion, an investor may wish to determine how much money he should money he should invest nowinvest now, at a fixed rate of interest, so , at a fixed rate of interest, so that he will that he will realize a certain sumrealize a certain sum at some at some future datefuture date..

This problem may be solved by This problem may be solved by expressingexpressing PP in terms ofin terms of AA..

1mt

rA P

m 1

mtr

A Pm

Present ValuePresent Value

Present value formula for compound interestPresent value formula for compound interest

1mt

rP A

m

1

mtr

P Am

ExamplesExamples

How much moneyHow much money should be should be depositeddeposited in a bank paying a in a bank paying a yearly interest rate of yearly interest rate of 6%6% compounded monthlycompounded monthly so that so that after after 33 years the years the accumulated amountaccumulated amount will be will be $20,000$20,000??

SolutionSolution Here, Here, AA = 20,000 = 20,000, , rr = 0.06 = 0.06, , mm = 12 = 12, and , and tt = 3 = 3.. Using the Using the present valuepresent value formula we get formula we get

(12)(3)

1

0.0620,000 1

12

16,713

mtr

P Am

(12)(3)

1

0.0620,000 1

12

16,713

mtr

P Am

Example 3, page 351

ExamplesExamples

Find the Find the present valuepresent value of of $49,158.60$49,158.60 duedue in in 55 years at an years at an interest rateinterest rate of of 10%10% per year compounded per year compounded quarterlyquarterly..

SolutionSolution Here, Here, AA = 49,158.60 = 49,158.60, , rr = 0.1 = 0.1, , mm = 4 = 4, and , and tt = 5 = 5.. Using the Using the present valuepresent value formula we get formula we get

(4)(5)

1

0.149,158.6 1

4

30,000

mtr

P Am

(4)(5)

1

0.149,158.6 1

4

30,000

mtr

P Am

Example 4, page 350

Continuous Compounding of InterestContinuous Compounding of Interest

One question arises on compound interest: One question arises on compound interest: ✦ What happens to the What happens to the accumulated amountaccumulated amount over a fixed over a fixed

period of time if the interest is period of time if the interest is compoundedcompounded more and more and more frequentlymore frequently??

We’ve seen that We’ve seen that the more oftenthe more often interest is interest is compoundedcompounded, , the largerthe larger the the accumulated amountaccumulated amount..

But does the But does the accumulated amountaccumulated amount approach aapproach a limitlimit when when interest is computed more and more frequently?interest is computed more and more frequently?


Recall that in the Recall that in the compound interestcompound interest formula formula

the the number of conversion periodsnumber of conversion periods is is mm.. So, we should let So, we should let mm get get larger and largerlarger and larger (approach (approach

infinity) and see what happens to the infinity) and see what happens to the accumulated accumulated amountamount AA..

But first, But first, for clarityfor clarity, lets , lets rewrite the equationrewrite the equation as follows: as follows:

1mt

rA P

m 1

mtr

A Pm

1

tmr

A Pm

1

tmr

A Pm


Letting Letting mm →→ , we find that, we find that

We can substitute We can substitute u u == m/r m/r (note that (note that uu →→ as as mm →→ ).). ThusThus

lim 1 lim 1

t tm m

m m

r rP P

m m

lim 1 lim 1

t tm m

m m

r rP P

m m

1 1lim 1 lim 1

t rtur u

u uP P

u u

1 1lim 1 lim 1

t rtur u

u uP P

u u


Now, you may recall thatNow, you may recall that

So, we can restate as follows:So, we can restate as follows:

Thus, as the Thus, as the frequencyfrequency with which interest is with which interest is compoundedcompounded increases without boundincreases without bound, the , the accumulated amountaccumulated amount approachesapproaches PePertrt..

1lim 1

u

ue

u

1lim 1

u

ue

u

1lim 1

rturt

uP Pe

u

1lim 1

rturt

uP Pe

u


Continuous Compound Interest FormulaContinuous Compound Interest Formula

A A == Pe Pertrt

wherewhere

PP == PrincipalPrincipal

rr == Annual interest rate compounded Annual interest rate compounded continuously.continuously.

tt == Time in years.Time in years.

AA == Accumulated amount at the end Accumulated amount at the end of of tt years. years.

ExamplesExamples

Find the Find the accumulated amountaccumulated amount after after 33 years if years if $1000$1000 is is invested at invested at 8%8% per year compounded per year compounded (a)(a) daily, and daily, and (b)(b) continuously. continuously.

SolutionSolution

a.a. Using the Using the compound interest compound interest formula with formula with PP = 1000 = 1000, , rr = 0.08 = 0.08, , mm = 365 = 365, and , and tt = 3 = 3, we find, we find

b.b. Using the Using the continuous compound interestcontinuous compound interest formula with formula with PP = 1000 = 1000, , rr = 0.08 = 0.08, and , and tt = 3 = 3, we find , we find

A A == Pe Pertrt = 1000= 1000ee(0.08)(3)(0.08)(3) ≈≈ 1271.25 1271.25

Note that both solutions are Note that both solutions are very similarvery similar..

(365)(3)0.08

1 1000 1 1271.22365

mtr

A Pm

(365)(3)0.08

1 1000 1 1271.22365

mtr

A Pm

Example 5, page 352

ExamplesExamples

How longHow long will it take will it take $10,000$10,000 to grow to to grow to $15, 000$15, 000 if the if the investment earns an investment earns an interestinterest raterate of of 12%12% per year per year compounded quarterlycompounded quarterly??

SolutionSolution Using the Using the compound interest compound interest formula with formula with AA = 15,000 = 15,000, ,

PP = 10,000 = 10,000, , rr = 0.12 = 0.12, and , and mm = 4 = 4, we obtain, we obtain

4

4

0.1215,000 10,000 1

4

15,000(1.03) 1.5

10,000

t

t

4

4

0.1215,000 10,000 1

4

15,000(1.03) 1.5

10,000

t

t

Example 7, page 354

ExamplesExamples

How longHow long will it take will it take $10,000$10,000 to grow to to grow to $15, 000$15, 000 if the if the investment earns an investment earns an interestinterest raterate of of 12%12% per year per year compounded quarterlycompounded quarterly??

SolutionSolution Taking Taking logarithmslogarithms on both sides gives on both sides gives

So, it will take approximately So, it will take approximately 3.43.4 yearsyears for the investment for the investment to grow from to grow from $10,000$10,000 to to $15,000$15,000..

4ln(1.03) ln1.5

4 ln1.03 ln1.5

ln1.5

4ln1.033.43

t

t

t

t

4ln(1.03) ln1.5

4 ln1.03 ln1.5

ln1.5

4ln1.033.43

t

t

t

t

Example 7, page 354

ExamplesExamples Find the Find the interest rateinterest rate needed for an needed for an investmentinvestment of of $10,000$10,000

to to grow to an amountgrow to an amount of of $18,000$18,000 in in 5 5 years if the interest is years if the interest is compounded monthlycompounded monthly..

SolutionSolution Using the Using the compound interest compound interest formula with formula with AA = 18,000 = 18,000, ,

PP = 10,000 = 10,000, , mm = 12 = 12, and , and tt = 5 = 5, we find, we find

(12)(5)

60

1

18,000 10,000 112

18,0001 1.8

12 10,000

mtr

A Pm

r

r

(12)(5)

60

1

18,000 10,000 112

18,0001 1.8

12 10,000

mtr

A Pm

r

r

Example 8, page 355



SolutionSolution Taking the Taking the 6060thth rootroot on both sides and solving for on both sides and solving for rr we get we get

60

60

60

60

1 1.812

1 1.812

1.8 112

12 1.8 1 0.009796

r

r

r

r

60

60

60

60

1 1.812

1 1.812

1.8 112

12 1.8 1 0.009796

r

r

r

r

Example 8, page 355



SolutionSolution Converting back into an Converting back into an exponential equationexponential equation, ,

andand

Thus, the Thus, the interest rateinterest rate needed is approximately needed is approximately 11.81%11.81% per year.per year.

0.009796112

1.009844

re

0.009796112

1.009844

re

1.009844 112

0.1181

r

r

1.009844 112

0.1181

r

r

Example 8, page 355

5.45.4Differentiation of the Exponential FunctionDifferentiation of the Exponential Function

1/212

,e 1/212

,e

1

x

y

– 1 1

1/212

,e 1/212

,e

2

( ) xf x e2

( ) xf x e

Rule 1Rule 1Derivative of the Exponential FunctionDerivative of the Exponential Function

The derivative of the exponential function with The derivative of the exponential function with base base ee is equal to the function itself: is equal to the function itself:

x xde e

dx x xd

e edx

ExamplesExamples

Find the Find the derivativederivative of the function of the function

SolutionSolution Using the Using the product ruleproduct rule gives gives

2( ) xf x x e 2( ) xf x x e

2 2 2

2

( )

(2 )

( 2)

x x x

x x

x

d d df x x e x e e x

dx dx dx

x e e x

xe x

2 2 2

2

( )

(2 )

( 2)

x x x

x x

x

d d df x x e x e e x

dx dx dx

x e e x

xe x

Example 1, page 362

ExamplesExamples


SolutionSolution Using the Using the general power rulegeneral power rule gives gives

3/2( ) 2tg t e 3/2( ) 2tg t e

1/2

1/2

1/2

3( ) 2 2

23

223

22

t t

t t

t t

dg t e e

dt

e e

e e

1/2

1/2

1/2

3( ) 2 2

23

223

22

t t

t t

t t

dg t e e

dt

e e

e e

Example 1, page 362

Rule 2Rule 2Chain Rule for Exponential FunctionsChain Rule for Exponential Functions

If If ff((xx)) is a differentiable function, then is a differentiable function, then

( ) ( ) ( )f x f xde e f x

dx ( ) ( ) ( )f x f xd

e e f xdx

ExamplesExamples


SolutionSolution

2( ) xf x e 2( ) xf x e

2

2

2

( ) 2

(2)

2

x

x

x

df x e x

dx

e

e

2

2

2

( ) 2

(2)

2

x

x

x

df x e x

dx

e

e

Example 2, page 363

ExamplesExamples


SolutionSolution

3xy e 3xy e

3

3

3

( 3 )

( 3)

3

x

x

x

dy de x

dx dx

e

e

3

3

3

( 3 )

( 3)

3

x

x

x

dy de x

dx dx

e

e

Example 2, page 363

ExamplesExamples


SolutionSolution

22( ) t tg t e 22( ) t tg t e

2

2

2 2

2

( ) 2

(4 1)

t t

t t

dg t e t t

dt

t e

2

2

2 2

2

( ) 2

(4 1)

t t

t t

dg t e t t

dt

t e

Example 2, page 363

ExamplesExamples


SolutionSolution

2 xy xe 2 xy xe

2 2

2 2

2 2

2 2

2

2 (1)

( 2)

2

(1 2 )

x x

x x

x x

x x

x

dy d dx e e x

dx dx dxd

x e x edx

xe e

xe e

e x

2 2

2 2

2 2

2 2

2

2 (1)

( 2)

2

(1 2 )

x x

x x

x x

x x

x

dy d dx e e x

dx dx dxd

x e x edx

xe e

xe e

e x

Example 3, page 363

ExamplesExamples


SolutionSolution

( )t

t t

eg t

e e

( )t

t t

eg t

e e

2

2

2 2

2

2

( )

1 1

2

t t t t t t

t t

t t t t t t

t t

t t

t t

t t

d de e e e e e

dt dtg te e

e e e e e e

e e

e e

e e

e e

2

2

2 2

2

2

( )

1 1

2

t t t t t t

t t

t t t t t t

t t

t t

t t

t t

d de e e e e e

dt dtg te e

e e e e e e

e e

e e

e e

e e

Example 4, page 364

ExamplesExamples

Find the Find the inflection pointsinflection points of the function of the function

SolutionSolution Find the Find the first and second derivativesfirst and second derivatives of of ff : :

Setting Setting ff″″ = 0= 0 gives gives ee–– xx22

= 0= 0, and , and 22xx22 – 1 = 0 – 1 = 0.. Since Since ee––

xx22

never equals zeronever equals zero for any real value of for any real value of xx, the only , the only candidates for inflectioncandidates for inflection pointspoints of of ff are are

Testing valuesTesting values around these numbers we conclude that around these numbers we conclude that

they are indeed inflection pointsthey are indeed inflection points..

2

( ) xf x e2

( ) xf x e

2

2 2

2 2

( ) 2

( ) ( 2 )( 2 ) 2

2 (2 1)

x

x x

x

f x xe

f x x xe e

e x

2

2 2

2 2

( ) 2

( ) ( 2 )( 2 ) 2

2 (2 1)

x

x x

x

f x xe

f x x xe e

e x

1 / 2x 1 / 2x

Example 6, page 364

ExamplesExamples

Find the Find the inflection pointsinflection points of the function of the function

SolutionSolution

2

( ) xf x e2

( ) xf x e

1/212

,e 1/212

,e

11

xx

yy

– – 11 1 1

1/212

,e 1/212

,e

2

( ) xf x e2

( ) xf x e

Example 6, page 364

5.55.5Differentiation of Logarithmic FunctionsDifferentiation of Logarithmic Functions

2 2 3( 1)( 4)y x x x 2 2 3( 1)( 4)y x x x 2 2 3

2 2 3

2

ln ln[ ( 1)( 4) ]

ln( ) ln( 1) ln( 4)

2 ln ln( 1) 3ln( 4)

y x x x

x x x

x x x

2 2 3

2 2 3

2

ln ln[ ( 1)( 4) ]

ln( ) ln( 1) ln( 4)

2 ln ln( 1) 3ln( 4)

y x x x

x x x

x x x

Rule 3Rule 3Derivative of the Natural LogarithmDerivative of the Natural Logarithm

The derivative of The derivative of lnln xx is is

1ln ( 0)

dx x

dx x

1ln ( 0)

dx x

dx x

ExamplesExamples


SolutionSolution

( ) lnf x x x( ) lnf x x x

( ) (ln ) ln ( )

1ln (1)

1 ln

d df x x x x x

dx dx

x xx

x

( ) (ln ) ln ( )

1ln (1)

1 ln

d df x x x x x

dx dx

x xx

x

Example 1, page 372

ExamplesExamples


SolutionSolution

ln( )

xg x

x

ln( )

xg x

x

2

2

2

(ln ) ln ( )( )

1ln (1)

1 ln

d dx x x x

dx dxg xx

x xx

xx

x

2

2

2

(ln ) ln ( )( )

1ln (1)

1 ln

d dx x x x

dx dxg xx

x xx

xx

x

Example 1, page 372

Rule 4Rule 4Chain Rule for Logarithmic FunctionsChain Rule for Logarithmic Functions

If If ff((xx)) is a differentiable function, then is a differentiable function, then

( )ln ( ) [ ( ) 0]

( )

d f xf x f x

dx f x

( )

ln ( ) [ ( ) 0]( )

d f xf x f x

dx f x

ExamplesExamples


SolutionSolution

2( ) ln( 1)f x x 2( ) ln( 1)f x x

2

2

2

1( )

12

1

dx

dxf xxx

x

2

2

2

1( )

12

1

dx

dxf xxx

x

Example 2, page 373

ExamplesExamples


SolutionSolution

2 3 6ln[( 1)( 2) ]y x x 2 3 6ln[( 1)( 2) ]y x x

2 3 6

2 3 6

2 3

ln[( 1)( 2) ]

ln( 1) ln( 2)

ln( 1) 6ln( 2)

y x x

x x

x x

2 3 6

2 3 6

2 3

ln[( 1)( 2) ]

ln( 1) ln( 2)

ln( 1) 6ln( 2)

y x x

x x

x x

2 3

2 3

2

2 3

2

2 3

( 1) ( 2)6

1 2

2 36

1 2

2 18

1 2

d dx xdy dx dx

dx x x

x x

x x

x x

x x

2 3

2 3

2

2 3

2

2 3

( 1) ( 2)6

1 2

2 36

1 2

2 18

1 2

d dx xdy dx dx

dx x x

x x

x x

x x

x x

Example 3, page 373

Logarithmic DifferentiationLogarithmic Differentiation

We have seen how finding derivatives of logarithmic We have seen how finding derivatives of logarithmic functions becomes easier when applying the functions becomes easier when applying the laws of laws of logarithmslogarithms..

These laws can also be used in a process called These laws can also be used in a process called logarithmic logarithmic differentiationdifferentiation to permit the differentiation of functions to permit the differentiation of functions that would bethat would be difficult to differentiate difficult to differentiate or evenor even not be not be differentiable differentiable through other means.through other means.

ExamplesExamples

Use Use logarithmic differentiationlogarithmic differentiation to find the to find the derivativederivative of of

SolutionSolution Take the Take the natural logarithmnatural logarithm of of both sidesboth sides of the equation: of the equation:

Use the Use the laws of logarithmslaws of logarithms to rewrite the equation: to rewrite the equation:

DifferentiateDifferentiate both sidesboth sides of the equation: of the equation:

2( 1)( 1)y x x x 2( 1)( 1)y x x x

2ln ln[ ( 1)( 1)]y x x x 2ln ln[ ( 1)( 1)]y x x x

2ln ln( ) ln( 1) ln( 1)y x x x 2ln ln( ) ln( 1) ln( 1)y x x x

2

2

ln ln( ) ln( 1) ln( 1)

1 1 2

1 1

d d d dy x x x

dx dx dx dxx

x x x

2

2

ln ln( ) ln( 1) ln( 1)

1 1 2

1 1

d d d dy x x x

dx dx dx dxx

x x x

Example 5, page 374

ExamplesExamples


SolutionSolution On the On the left sideleft side, note that , note that yy is a function ofis a function of xx, therefore:, therefore:

2( 1)( 1)y x x x 2( 1)( 1)y x x x

( )

ln ln[ ( )]

ln ln[ ( )]

( )

( )

y f x

y f x

d dy f x

dx dxf x

f x

y

y

( )

ln ln[ ( )]

ln ln[ ( )]

( )

( )

y f x

y f x

d dy f x

dx dxf x

f x

y

y

Example 5, page 374

ExamplesExamples


SolutionSolution Thus, we have:Thus, we have:

2( 1)( 1)y x x x 2( 1)( 1)y x x x

2

2

1 1 2ln

1 11 1 2

1 1

d xy

dx x x xy x

y x x x

2

2

1 1 2ln

1 11 1 2

1 1

d xy

dx x x xy x

y x x x

Example 5, page 374

ExamplesExamples


SolutionSolution Finally, Finally, solving forsolving for yy′′ we get we get::

2( 1)( 1)y x x x 2( 1)( 1)y x x x

2

22

1 1 2

1 1

1 1 2( 1)( 1)

1 1

xy y

x x x

xx x x

x x x

2

22

1 1 2

1 1

1 1 2( 1)( 1)

1 1

xy y

x x x

xx x x

x x x

Example 5, page 374

Logarithmic DifferentiationLogarithmic Differentiation

To find To find dy/dxdy/dx by logarithmic differentiation: by logarithmic differentiation:

1.1. Take the Take the natural logarithmnatural logarithm on on both sidesboth sides of the of the equation and use the equation and use the properties of logarithmsproperties of logarithms to write any “complicated expression” as a sum to write any “complicated expression” as a sum of of simpler termssimpler terms..

2.2. Differentiate both sidesDifferentiate both sides of the equation with of the equation with respect to respect to xx..

3.3. SolveSolve the resulting equation for the resulting equation for dy/dxdy/dx..

ExamplesExamples


SolutionSolution

1.1. Take the Take the natural logarithmnatural logarithm of of both sidesboth sides of the equation of the equation and use the and use the laws of logarithmslaws of logarithms to to rewriterewrite the equation: the equation:

2 2 3( 1)( 4)y x x x 2 2 3( 1)( 4)y x x x

2 2 3

2 2 3

2

ln ln[ ( 1)( 4) ]

ln( ) ln( 1) ln( 4)

2 ln ln( 1) 3ln( 4)

y x x x

x x x

x x x

2 2 3

2 2 3

2

ln ln[ ( 1)( 4) ]

ln( ) ln( 1) ln( 4)

2 ln ln( 1) 3ln( 4)

y x x x

x x x

x x x

Example 6, page 375

ExamplesExamples


SolutionSolution

2.2. DifferentiateDifferentiate both sidesboth sides of the equation: of the equation:

2 2 3( 1)( 4)y x x x 2 2 3( 1)( 4)y x x x

2

2

2

ln 2 ln ln( 1) 3 ln( 4)

1 1 22 3

1 42 1 6

1 4

d d d dy x x x

dx dx dx dxx

x x xx

x x x

2

2

2

ln 2 ln ln( 1) 3 ln( 4)

1 1 22 3

1 42 1 6

1 4

d d d dy x x x

dx dx dx dxx

x x xx

x x x

Example 6, page 375

ExamplesExamples


SolutionSolution

3.3. SolveSolve for for dy/dxdy/dx::

2

2

2 1 6ln

1 42 1 6

1 4

d xy

dx x x xy x

y x x x

2

2

2 1 6ln

1 42 1 6

1 4

d xy

dx x x xy x

y x x x

2

2 2 32

2 1 6

1 4

2 1 6( 1)( 4)

1 4

xy y

x x x

xx x x

x x x

2

2 2 32

2 1 6

1 4

2 1 6( 1)( 4)

1 4

xy y

x x x

xx x x

x x x

2 2 3( 1)( 4)y x x x 2 2 3( 1)( 4)y x x x

Example 6, page 375

ExamplesExamples


SolutionSolution

1.1. Take the Take the natural logarithmnatural logarithm of of both sidesboth sides of the equation of the equation and use the and use the laws of logarithmslaws of logarithms to to rewriterewrite the equation: the equation:

( ) ( 0)xf x x x ( ) ( 0)xf x x x

ln ( ) ln

ln

xf x x

x x

ln ( ) ln

ln

xf x x

x x

Example 7, page 376

ExamplesExamples


SolutionSolution

2.2. DifferentiateDifferentiate both sidesboth sides of the equation: of the equation:

ln ( ) (ln ) ln ( )

1ln (1)

1 ln

d d df x x x x x

dx dx dx

x xx

x

ln ( ) (ln ) ln ( )

1ln (1)

1 ln

d d df x x x x x

dx dx dx

x xx

x

( ) ( 0)xf x x x ( ) ( 0)xf x x x

Example 7, page 376

ExamplesExamples


SolutionSolution

3.3. SolveSolve for for dy/dxdy/dx::

ln ( ) 1 ln

( )1 ln

( )

df x x

dxf x

xf x

ln ( ) 1 ln

( )1 ln

( )

df x x

dxf x

xf x

( ) ( )(1 ln )

(1 ln )x

f x f x x

x x

( ) ( )(1 ln )

(1 ln )x

f x f x x

x x

( ) ( 0)xf x x x ( ) ( 0)xf x x x

Example 7, page 376

5.65.6Exponential Functions as Mathematical ModelsExponential Functions as Mathematical Models

1.1. Growth of bacteriaGrowth of bacteria

2.2. Radioactive decayRadioactive decay

3.3. Assembly timeAssembly time

Applied Example:Applied Example: Growth of Bacteria Growth of Bacteria

Under a laboratory, the Under a laboratory, the number of bacterianumber of bacteria in a culture in a culture grows according togrows according to

where where QQ00 denotes the number of denotes the number of bacteria initially presentbacteria initially present

in the culture, in the culture, kk is a is a constantconstant determined by the determined by the strain of strain of bacteriabacteria under consideration, and under consideration, and tt is the is the elapsed timeelapsed time measured in hours.measured in hours.

Suppose Suppose 10,000 10,000 bacteria are bacteria are present initiallypresent initially in the culture in the culture and and 60,00060,000 present present two hours latertwo hours later..

a.a. How many bacteriaHow many bacteria will there be in the culture at the will there be in the culture at the end of end of four hoursfour hours??

b.b. What is the What is the rate of growthrate of growth of the population after of the population after four four hourshours??

0( ) ktQ t Q e 0( ) ktQ t Q e

Applied Example 1, page 380


SolutionSolution

a.a. We are given that We are given that QQ(0) = (0) = QQ00 = 10,000 = 10,000, so , so QQ((tt) = 10,000) = 10,000eektkt..

At At tt = 2 = 2 there are there are 60,000 60,000 bacteria, so bacteria, so QQ(2) = 60,000(2) = 60,000, thus:, thus:

Taking the Taking the natural logarithmnatural logarithm on on both sidesboth sides we get: we get:

So, the So, the number of bacteria presentnumber of bacteria present at any time at any time tt is given by: is given by:

02

2

( )60,000 10,000

6

kt

k

k

Q t Q ee

e

02

2

( )60,000 10,000

6

kt

k

k

Q t Q ee

e

2ln ln 6

2 ln 6

0.8959

ke

k

k

2ln ln 6

2 ln 6

0.8959

ke

k

k

0.8959( ) 10,000 tQ t e 0.8959( ) 10,000 tQ t eApplied Example 1, page 380


SolutionSolution

a.a. At the end of At the end of four hoursfour hours ( (tt = 4 = 4), there will be), there will be

or or 360,029360,029 bacteriabacteria..

0.8959(4)(4) 10,000

360,029

Q e

0.8959(4)(4) 10,000

360,029

Q e



SolutionSolution

b.b. The The rate of growthrate of growth of the bacteria at any time of the bacteria at any time tt is given by is given by

Using the result from part Using the result from part (a)(a), we find that the , we find that the rate of rate of bacterial growthbacterial growth at the end of at the end of four hoursfour hours is is

or approximately or approximately 322,550322,550 bacteria per hourbacteria per hour..

( ) ( )Q t kQ t ( ) ( )Q t kQ t

(4) (4)

(0.8959)(360,029)

322,550

Q kQ

(4) (4)

(0.8959)(360,029)

322,550

Q kQ


Applied Example:Applied Example: Radioactive Decay Radioactive Decay

Radioactive substances Radioactive substances decay exponentiallydecay exponentially.. For example, the amount of For example, the amount of radiumradium present at any time present at any time tt

obeys the law obeys the law

where where QQ00 is the is the initial amountinitial amount present and present and kk is a suitable is a suitable

positive positive constantconstant.. The The half-life half-life of a radioactive substance is the time of a radioactive substance is the time

required for a given amount to be required for a given amount to be reduced by one-halfreduced by one-half.. The The half-life half-life ofof radium radium is approximately is approximately 16001600 years. years. Suppose initially there are Suppose initially there are 200200 milligrams of pure radium. milligrams of pure radium.

a.a. Find the amount left after Find the amount left after tt years. years.

b.b. What is the amount after What is the amount after 800800 years? years?

0( ) (0 )ktQ t Q e t 0( ) (0 )ktQ t Q e t



SolutionSolution

a.a. The The initial amountinitial amount is is 200200 milligrams, so milligrams, so QQ(0) = (0) = QQ00 = 200 = 200, so , so

QQ((tt) = 200) = 200ee––ktkt

The The half-life of radiumhalf-life of radium is is 16001600 years, so years, so QQ(1600) = 100(1600) = 100, thus, thus1600

1600

100 200

1

2

k

k

e

e

1600

1600

100 200

1

2

k

k

e

e



SolutionSolution

a.a. Taking the Taking the natural logarithmnatural logarithm on on both sidesboth sides yields: yields:

Therefore, the Therefore, the amount of radium leftamount of radium left after after tt years is: years is:

1600 1ln ln

21

1600 ln ln21

1600 ln21 1

ln 0.00043321600 2

ke

k e

k

k

1600 1ln ln

21

1600 ln ln21

1600 ln21 1

ln 0.00043321600 2

ke

k e

k

k

0.0004332( ) 200 tQ t e 0.0004332( ) 200 tQ t eApplied Example 2, page 382


SolutionSolution

b.b. In particular, the In particular, the amount of radium amount of radium left left afterafter 800800 years is: years is:

or approximately or approximately 141141 milligrams. milligrams.

0.0004332(800)(800) 200

141.42

Q e

0.0004332(800)(800) 200

141.42

Q e


Applied Example:Applied Example: Assembly Time Assembly Time The Camera Division of Eastman Optical produces a The Camera Division of Eastman Optical produces a single single

lens reflexlens reflex cameracamera.. Eastman’s Eastman’s training departmenttraining department determines that after determines that after

completing the basic training program, a new, previously completing the basic training program, a new, previously inexperienced employeeinexperienced employee will be able to assemble will be able to assemble

model F cameras per day,model F cameras per day, t t monthsmonths after the employee after the employee starts work on the assembly line.starts work on the assembly line.a.a. How manyHow many model F cameras can a model F cameras can a new employeenew employee assemble assemble

per day per day after basic trainingafter basic training??b.b. How manyHow many model F cameras can an employee with model F cameras can an employee with one one

month of experiencemonth of experience assemble per day? assemble per day?c.c. How manyHow many model F cameras can the model F cameras can the average experiencedaverage experienced

employee assemble per day?employee assemble per day?

0.5( ) 50 30 tQ t e 0.5( ) 50 30 tQ t e


Applied Example:Applied Example: Assembly Time Assembly TimeSolutionSolutiona.a. The number of model F cameras a The number of model F cameras a new employeenew employee can can

assembleassemble is given by is given by

b.b. The number of model F cameras that an employee with The number of model F cameras that an employee with 11, , 22, and , and 66 months of experiencemonths of experience can can assemble per dayassemble per day is is given bygiven by

or about or about 3232 cameras per day. cameras per day.

c.c. As As tt increasesincreases without bound, without bound, QQ((tt)) approachesapproaches 5050. . Hence, the Hence, the average experienced employeeaverage experienced employee can be expected can be expected to to assembleassemble 5050 model F cameras per day. model F cameras per day.

(0) 50 30 20Q (0) 50 30 20Q

0.5(1)(1) 50 30 31.80Q e 0.5(1)(1) 50 30 31.80Q e


End of End of Chapter Chapter

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5 Exponential Functions Logarithmic Functions Compound Interest Differentiation of Exponential Functions Exponential Functions as Mathematical