3 Exponential Functions Logarithmic Functions Exponential Functions as Mathematical Models Exponential and Logarithmic Functions

33

Exponential FunctionsExponential Functions

Logarithmic FunctionsLogarithmic Functions

Exponential Functions as Exponential Functions as Mathematical ModelsMathematical Models

Exponential and Logarithmic FunctionsExponential and Logarithmic Functions

3.13.1Exponential 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 (–(– , , )). . The 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 (–(– , , )). . The 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 = = ..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))

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

ExamplesExamples


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

xx

yy

– – 2 2 22

44

22

ff((xx) = 2) = 2xx

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))

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

ExamplesExamples


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

xx

yy

– – 2 2 22

44

22

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

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

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

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

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

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

3.23.2Logarithmic 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. .

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

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

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

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

ExamplesExamples


log81log81 4log3

4log3

4(0.4771)

1.9084

4log3

4log3

4(0.4771)

1.9084

ExamplesExamples


log50log50 log5 10

log5 log10

0.6990 1

1.6990

log5 10

log5 log10

0.6990 1

1.6990

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

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

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

ExamplesExamples

Use the Use the properties of logarithmsproperties of logarithms to to solve solve the equation forthe equation for xx::

3 3log ( 1) log ( 1) 1x x 3 3log ( 1) log ( 1) 1x x

3

1log 1

1

x

x

3

1log 1

1

x

x

113 3

1

x

x

11

3 31

x

x

1 3( 1)x x 1 3( 1)x x

1 3 3x x 1 3 3x x

4 2x4 2x

2x 2x

Law 2Law 2

Definition of Definition of logarithmslogarithms

ExamplesExamples

Use the Use the properties of logarithmsproperties of logarithms to to solve solve the equation forthe equation for xx::

log log(2 1) log6x x log log(2 1) log6x x

log log(2 1) log6 0x x log log(2 1) log6 0x x

(2 1)log 0

6

x x

(2 1)log 0

6

x x

0(2 1)10 1

6

x x 0(2 1)

10 16

x x

(2 1) 6x x (2 1) 6x x 22 6 0x x 22 6 0x x

(2 3)( 2) 0x x (2 3)( 2) 0x x

2x 2x

Laws 1 and 2Laws 1 and 2

Definition of Definition of logarithmslogarithms

3

2log

x

x

is out of

the domain of , so it is discarded.

3

2log

x

x

is out of

the domain of , so it is discarded.

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::

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

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

3.33.3Exponential 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

In a laboratory, the In a laboratory, the number of bacterianumber of bacteria in a culture grows in a culture grows according toaccording 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..

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

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


SolutionSolution 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 e


SolutionSolution 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

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

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, thus

1600

1600

100 200

1

2

k

k

e

e

1600

1600

100 200

1

2

k

k

e

e


SolutionSolution

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


SolutionSolution

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

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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3 Exponential Functions Logarithmic Functions Exponential Functions as Mathematical Models Exponential and Logarithmic Functions