Mathematics for Economists 3: Exponential and Logarithmic Functions 10/8/2017 Econ 506, by Dr. M. Zainal 1 Mathematics for Economists Department of Economics Econ 506 Dr. Mohammad

Chapter 3: Exponential and Logarithmic Functions

10/8/2017

Econ 506, by Dr. M. Zainal 1

Mathematics for Economists

Department of Economics

Dr. Mohammad ZainalEcon 506

Chapter 3Exponential and Logarithmic

Functions

3

•Exponential Functions

•Logarithmic Functions

•Exponential Functions as Mathematical Models

Econ 506, by Dr. M. Zainal Ch 3 - 2

Chapter Outline


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3.1Exponential Functions


x

y

4

2

f(x) = (1/2)x

f(x) = 2x

– 2 2

Exponential Function• The function defined by

is called an exponential function with base b and exponentx.

• The domain of f is the set of all real numbers.

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



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Example• The exponential function with base 2 is the function

with domain (–, ).

• The values of f(x) for selected values of x 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


Example• The exponential function with base 2 is the function

with domain (–, ).

• The values of f(x) for selected values of x follow:

( 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


( ) 2xf x ( ) 2xf x


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• Let a and b be positive numbers and let x and y be real numbers. Then,


Laws of Exponents

• Let f(x) = 22x – 1. Find the value of x for which f(x) = 16.

Solution

• We want to solve the equation

22x – 1 = 16 = 24

• But this equation holds if and only if

2x – 1 = 4

giving x = .5

2


Examples


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• Sketch the graph of the exponential function f(x) = 2x.

Solution

• First, recall that the domain of this function is the set of real numbers.

• Next, putting x = 0 gives y = 20 = 1, which is the y-intercept.

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


Examples


Solution

• Now, consider a few values for x:

• Note that 2x approaches zero as x decreases without bound:• There is a horizontal asymptote at y = 0.

• Furthermore, 2x increases without bound when x increases without bound.

• Thus, the range of f is the interval (0, ).

x – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5

y 1/32 1/16 1/8 1/4 1/2 1 2 4 8 16 32


Examples


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Solution

• Finally, sketch the graph:

x

y

– 2 2

4

2

f(x) = 2x


Examples

• Sketch the graph of the exponential function f(x) = (1/2)x.

Solution

• First, recall again that the domain of this function is the set of realnumbers.

• Next, putting x = 0 gives y = (1/2)0 = 1, which is the y-intercept.

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


Examples


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Solution

• Now, consider a few values for x:

• Note that (1/2)x increases without bound when x decreases without bound.

• Furthermore, (1/2)x approaches zero as x increases without bound: there is a horizontal asymptote at y = 0.

• As before, the range of f is the interval (0, ).


Examples

x – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5

y 1/32 1/16 1/8 1/4 1/2 1 2 4 8 16 32


Solution

• Finally, sketch the graph:

x

y

– 2 2

4

2

f(x) = (1/2)x


Examples


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Examples• Sketch the graph of the exponential function f(x) = (1/2)x.

Solution

• Note the symmetry between the two functions:

x

y

4

2

f(x) = (1/2)x

f(x) = 2x


– 2 2

Properties of Exponential Functions• The exponential function y = bx (b > 0, b ≠ 1) has the

following properties:1. Its domain is (–, ).2. Its range is (0, ).3. Its graph passes through the point (0, 1)4. It is continuous on (–, ).5. It is increasing on (–, ) if b > 1 and decreasing on (–, )

if b < 1.



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The Base e• Exponential functions to the base e, where e is an irrational

number whose value is 2.7182818…, play an important role in both theoretical and applied problems.

• It can be shown that

1lim 1

m

me

m

1lim 1

m

me

m


Examples• Sketch the graph of the exponential function f(x) = ex.

Solution

• Since ex > 0 it follows that the graph of y = ex is similar to the graph of y =2x.

• Consider a few values for x:

x – 3 – 2 – 1 0 1 2 3

y 0.05 0.14 0.37 1 2.72 7.39 20.09



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5

3

1

Examples• Sketch the graph of the exponential function f(x) = ex.

Solution

• Sketching the graph:

x

y

– 3 – 1 1 3

f(x) = ex


Examples• Sketch the graph of the exponential function f(x) = e–x.

Solution

• Since e–x > 0 it follows that 0 < 1/e < 1 and so

f(x) = e–x = 1/ex = (1/e)x is an exponential function with base less than1.

• Therefore, it has a graph similar to that of y = (1/2)x.

• Consider a few values for x:

x – 3 – 2 – 1 0 1 2 3

y 20.09 7.39 2.72 1 0.37 0.14 0.05



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5

3

1

Examples• Sketch the graph of the exponential function f(x) = e–x.

Solution

• Sketching the graph:

x

y

– 3 – 1 1 3

f(x) = e–x


3.2Logarithmic Functions

1

x

y

1

y = ex

y = ln x

y = x



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Logarithms

• We’ve discussed exponential equations of the form

y = bx (b > 0, b ≠ 1)

• But what about solving the same equation for y?

• You may recall that y is called the logarithm of x to the base b, and is denoted logbx.

• Logarithm of x to the base b

y = logbx if and only if x = by (x > 0)


Examples• Solve log3x = 4 for x:

Solution

• By definition, log3x = 4 implies x = 34 = 81.


• Solve log164 = x for x:

Solution

• log164 = x is equivalent to 4 = 16x = (42)x = 42x, or 41 = 42x,

from which we deduce that

2 1

1

2

x

x

2 1

1

2

x

x


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Examples• Solve logx8 = 3 for x:

Solution

• By definition, we see that logx8 = 3 is equivalent to

3 38 2

2

x

x

3 38 2

2

x

x


Logarithmic Notation

log x = log10 x Common logarithm

ln x = loge x Natural logarithm



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Laws of Logarithms

• If m and n are positive numbers, then

1.

2.

3.

4.

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


Examples• Given that log 2 ≈ 0.3010, log 3 ≈ 0.4771, and log 5 ≈ 0.6990, use the laws

of logarithms to find

log15log15 log 3 5

log 3 log 5

0.4771 0.6990

1.1761

log 3 5

log 3 log 5

0.4771 0.6990

1.1761


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

log81log81 4log 3

4log 3

4(0.4771)

1.9084

4log 3

4log 3

4(0.4771)

1.9084


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Examples• Given that log 2 ≈ 0.3010, log 3 ≈ 0.4771, and log 5 ≈ 0.6990, use the laws

of logarithms to findlog 7.5log 7.5 log(15 / 2)

log(3 5 / 2)

log 3 log 5 log 2

0.4771 0.6990 0.3010

0.8751

log(15 / 2)

log(3 5 / 2)

log 3 log 5 log 2

0.4771 0.6990 0.3010

0.8751


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

log50log50 log 5 10

log 5 log10

0.6990 1

1.6990

log 5 10

log 5 log10

0.6990 1

1.6990

Examples• Expand and simplify 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


• Expand and simplify the expression:

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


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Examples• Expand and simplify the expression:

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


Examples• Use the properties of logarithms to solve the equation for x:

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 2

Definition of logarithms



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Examples• Use the properties of logarithms to solve the equation for x:

log log(2 1) log 6x x log log(2 1) log 6x x

log log(2 1) log 6 0x x log log(2 1) log 6 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

Laws 1 and 2

Definition of logarithms

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.


2x 2x

Logarithmic Function• The function defined by

is called the logarithmic function with base b.

• The domain of f is the set of all positive numbers.

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



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Properties of Logarithmic Functions

• The logarithmic function y = logbx (b > 0, b ≠ 1)

has the following properties:1. Its domain is (0, ).2. Its range is (–, ).3. Its graph passes through the point (1, 0).4. It is continuous on (0, ).5. It is increasing on (0, ) if b > 1

and decreasing on (0, ) if b < 1.


Example• Sketch the graph of the function y = ln x.

Solution

• We first sketch the graph of y = ex.

1

x

y y = ex

y = ln x

y = x• The required graph is

the mirror image of the graph of y = ex with respect to the line y = x


1


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Properties Relating Exponential and Logarithmic Functions

• Properties relating ex and ln x:

eln x = x (x > 0)

ln ex = x (for any real number x)


Examples• Solve the equation 2ex + 2 = 5.

Solution

• Divide both sides of the equation by 2 to obtain:

• Take the natural logarithm of each side of the equation and solve:

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



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

Solution

• Add – 3 to both sides of the equation and then divide both sides of the equation by 5 to obtain:

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


3.3Exponential Functions as Mathematical Models

1. Growth of bacteria2. Radioactive decay3. Assembly time



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Applied Example: Growth of Bacteria• In a laboratory, the number of bacteria in a culture grows according to

where Q0 denotes the number of bacteria initially present in the culture, k is a constant determined by the strain of bacteria under consideration, and t is the elapsed time measured in hours.

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

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

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


Applied Example: Growth of BacteriaSolution

• We are given that Q(0) = Q0 = 10,000, so Q(t) = 10,000ekt.

• At t = 2 there are 60,000 bacteria, so Q(2) = 60,000, thus:

• Taking the natural logarithm on both sides we get:

• So, the number of bacteria present at any time t 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



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Applied Example: Growth of BacteriaSolution

• At the end of four hours (t = 4), there will be

or 360,029 bacteria.

0.8959(4)(4) 10,000

360,029

Q e

0.8959(4)(4) 10,000

360,029

Q e


Applied Example: Radioactive Decay• Radioactive substances decay exponentially.

• For example, the amount of radium present at any time t obeys the law

where Q0 is the initial amount present and k is a suitable positive constant.

• The half-life of a radioactive substance is the time required for a given amount to be reduced by one-half.

• The half-life of radium is approximately 1600 years.

• Suppose initially there are 200 milligrams of pure radium.a. Find the amount left after t years.b. What is the amount after 800 years?

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



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Applied Example: Radioactive DecaySolution

a. Find the amount left after t years.

The initial amount is 200 milligrams, so Q(0) = Q0 = 200, so

Q(t) = 200e–kt

The half-life of radium is 1600 years, so Q(1600) = 100, thus

1600

1600

100 200

1

2

k

k

e

e

1600

1600

100 200

1

2

k

k

e

e



a. Find the amount left after t years. Taking the natural logarithm on both sides yields:

Therefore, the amount of radium left after t 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



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b. What is the amount after 800 years?

In particular, the amount of radium left after 800 years is:

or approximately 141 milligrams.

0.0004332(800)(800) 200

141.42

Q e

0.0004332(800)(800) 200

141.42

Q e


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

camera.

• Eastman’s training department determines that after completing the basic training program, a new, previously inexperienced employee will be able to assemble

model F cameras per day, t months after the employee starts work on the assembly line.

a. How many model F cameras can a new employee assemble per day after basictraining?

b. How many model F cameras can an employee with one month of experienceassemble per day?

c. How many model F cameras can the average experienced employee assemble per day?

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



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Applied Example: Assembly TimeSolution

a. The number of model F cameras a new employee can assemble is given by

b. The number of model F cameras that an employee with 1, 2, and 6months of experience can assemble per day is given by

or about 32 cameras per day.

c. As t increases without bound, Q(t) approaches 50.

Hence, the average experienced employee can be expected to assemble50 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 CHAPTER


Documents

Mathematics for Economists 3: Exponential and Logarithmic Functions 10/8/2017 Econ 506, by Dr. M. Zainal 1 Mathematics for Economists Department of Economics Econ 506 Dr. Mohammad