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Page 1: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-1Copyright © 2005 Pearson Education, Inc.

Page 2: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-2Copyright © 2005 Pearson Education, Inc.

Exponential and Logarithmic Functions

Chapter 5

Page 3: Slide 4-1 Copyright © 2005 Pearson Education, Inc

5.1Exponential Functions

and Graphs

· Graph exponential equations and functions.· Solve applied problems involving exponential

functions and their graphs.

Page 4: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-4Copyright © 2005 Pearson Education, Inc.

Exponential Function

The function f(x) = ax, where x is a real number, a > 0 and a 1, is called the exponential function, base a.

The base needs to be positive in order to avoid the complex numbers that would occur by taking even roots of negative numbers.

Examples:

1( ) 3 ( ) ( ) (4.23)

3

xx xf x f x f x

Page 5: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-5Copyright © 2005 Pearson Education, Inc.

Graphing Exponential Functions

To graph an exponential function, follow the steps listed:

1. Compute some function values and list the results

in a table.

2. Plot the points and connect them with a smooth curve. Be sure to plot enough points to

determine how steeply the curve rises.

Page 6: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-6Copyright © 2005 Pearson Education, Inc.

Example

Graph the exponential function y = f(x) = 3x.

(3, 1/27)1/273

(2, 1/9)1/92

(1, 1/3)1/31

(3, 27)273

9

3

1

y = f(x) = 3x

(2, 9)2

(1, 3)1

(0, 1)0

(x, y)x

Page 7: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-7Copyright © 2005 Pearson Education, Inc.

Example

Graph the exponential function . 1

( )3

x

y f x

(3, 1/27)1/273

(2, 1/9)1/92

(1, 1/3)1/31

(3, 27)273

9

3

1

(2, 9)2

(1, 3)1

(0, 1)0

(x, y)x 1( )

3

x

y f x

Page 8: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-8Copyright © 2005 Pearson Education, Inc.

Example

Graph y = 3x + 2.

The graph is the graph of y = 3x shifted to left 2 units.

2433

812

271

90

3

1

1/3

y

1

2

3

x

Page 9: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-9Copyright © 2005 Pearson Education, Inc.

Example

Graph y = 4 3x

The graph is a reflection of the graph of y = 3x across the x-axis, followed by a reflection across the y-axis and then a shift up of 4 units.

3.963

3.882

3.671

30

1

5

23

y

1

2

3

x

Page 10: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-10Copyright © 2005 Pearson Education, Inc.

The Number e

e 2.7182818284…

Find each value of ex, to four decimal places, using the ex key on a calculator.

a) e4 b) e0.25

c) e2 d) e1

a) 54.5982 b) 0.7788 c) 7.3891 d) 0.3679

Page 11: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-11Copyright © 2005 Pearson Education, Inc.

Graphs of Exponential Functions, Base e

Graph f(x) = ex.

7.3892

2.7181

1

0.368

0.135

f(x)

0

1

2

x

Page 12: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-12Copyright © 2005 Pearson Education, Inc.

Example

Graph f(x) = 2 e3x.

1.992

1.951

1

18.09

401.43

f(x)

0

1

2

x

Page 13: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-13Copyright © 2005 Pearson Education, Inc.

Example

Graph f(x) = ex+2.

0.1354

0.3683

20.0861

7.389

2.718

1

f(x)

0

1

2

x

Page 14: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Logarithmic Functions and Graphs

· Graph logarithmic functions.· Convert between exponential and logarithmic

equations.· Find common and natural logarithms using a

calculator.

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Slide 4-15Copyright © 2005 Pearson Education, Inc.

Logarithmic Functions

These functions are inverses of exponential functions.

Graph: x = 3y.

1. Choose values for y.

2. Compute values for x.

3. Plot the points and connect them with a smooth curve.

* Note that the curve does not touch or cross the y-axis.

Page 16: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-16Copyright © 2005 Pearson Education, Inc.

Logarithmic Functions continued

(1/27, 3)31/27

(1/9, 2)21/9

(1/3, 1)11/3

2

1

0

y

(9, 2)9

(3, 1)3

(1, 0)1

(x, y)x = 3y

Graph: x = 3y

Page 17: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-17Copyright © 2005 Pearson Education, Inc.

Logarithmic Function, Base a

We define y = loga x as that number y such that x = ay, where x > 0 and a is a positive constant other than 1.

We read loga x as “the logarithm, base a, of x.”

Page 18: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-18Copyright © 2005 Pearson Education, Inc.

Finding Certain Logarithms

Find each of the following logarithms. a) log2 16 b) log10 1000 c) log16 4 d) log10 0.001

a) The exponent to which we raise 2 to obtain 16 is 4; thus log2 16 = 4.b) The exponent to which we raise 10 to obtain 1000 is 3;

thus log10 1000 = 3.c) The exponent we raise 16 to get 4 is ½, so log16 4 = ½.d) We have The exponent to which we raise 10

to get 0.001 is 3, so log10 0.001 = 3.

33

1 110 .

1000 10

Page 19: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-19Copyright © 2005 Pearson Education, Inc.

Logarithms

loga 1 = 0 and loga a = 1, for any logarithmic base a.

Convert each of the following to a logarithmic equation.

a) 25 = 5x b) ew = 30

log5 25 = x loge 30 = w

log A logarithm is an exponent!ayyx x a

Page 20: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-20Copyright © 2005 Pearson Education, Inc.

Example

Convert each of the following to an exponential equation.

a) log7 343 = 3 log7 343 = 3 73 = 343

b) logb R = 12

logb R = 12 b12 = R

The logarithm is the exponent.

The base remains the same.

Page 21: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-21Copyright © 2005 Pearson Education, Inc.

Example

Find each of the following common logarithms on a calculator. Round to four decimal places.

a) log 723,456

b) log 0.0000245

c) log (4)

Does not existERR: nonreal anslog (4)

4.61084.610833916log 0.0000245

5.85945.859412123log 723,456

RoundedReadoutFunction Value

Page 22: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-22Copyright © 2005 Pearson Education, Inc.

Natural Logarithms

Logarithms, base e, are called natural logarithms. The abbreviation “ln” is generally used for natural logarithms. Thus,

ln x means loge x.

ln 1 = 0 and ln e = 1, for the logarithmic base e.

Page 23: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-23Copyright © 2005 Pearson Education, Inc.

Example

Find each of the following natural logarithms on a calculator. Round to four decimal places.

a) ln 723,456

b) ln 0.0000245

c) ln (4)

Does not existERR: nonreal ansln (4)

10.616810.61683744ln 0.0000245

13.491813.49179501ln 723,456

RoundedReadoutFunction Value

Page 24: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-24Copyright © 2005 Pearson Education, Inc.

Changing Logarithmic Bases

The Change-of-Base Formula

For any logarithmic bases a and b, and any positive number M,

loglog .

loga

ba

MM

b

Page 25: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-25Copyright © 2005 Pearson Education, Inc.

Example

Find log6 8 using common logarithms.

Solution: First, we let a = 10, b = 6, and M = 8. Then we substitute into the change-of-base formula:

10

16

0

loglog

l

8

og

1.1606

68

Page 26: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-26Copyright © 2005 Pearson Education, Inc.

Example

We can also use base e for a conversion.

Find log6 8 using natural logarithms.

Solution: Substituting e for a, 6 for b and 8 for M, we have

6 6

loglog

log

ln81.1606

ln 6

88 e

e

Page 27: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-27Copyright © 2005 Pearson Education, Inc.

Graphs of Logarithmic Functions

Graph: y = f(x) = log6 x. Select y. Compute x.

21/36

11/6

3216

236

16

01

yx,or 6y

Page 28: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-28Copyright © 2005 Pearson Education, Inc.

Example

Graph each of the following. Describe how each graph can be obtained from the graph of y = ln x. Give the domain and the vertical asymptote of each function.

a) f(x) = ln (x 2) b) f(x) = 2 ln x c) f(x) = |ln (x + 1)|

14

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Slide 4-29Copyright © 2005 Pearson Education, Inc.

Graph f(x) = ln (x 2)

The graph is a shift 2 units right. The domain is the set of all real numbers greater than 2. The line x = 2 is the vertical asymptote.

1.0995

0.6934

03

0.6932.5

1.3862.25

f(x)x

Page 30: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-30Copyright © 2005 Pearson Education, Inc.

Graph f(x) = 2 ln x

The graph is a vertical shrinking, followed by a reflection across the x-axis, and then a translation up 2 units. The domain is the set of all positive real numbers. The y-axis is the vertical asymptote.

14

1.5985

1.7253

21

2.1730.5

2.5760.1

f(x)x

Page 31: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-31Copyright © 2005 Pearson Education, Inc.

Graph f(x) = |ln (x + 1)|

The graph is a translation 1 unit to the left. Then the absolute value has the effect of reflecting negative outputs across the x-axis. The domain is the set of all real numbers greater than 1. The line x = 1 is the vertical asymptote.

1.9466

1.3863

0.6931

00

0.6930.5

f(x)x

Page 32: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-32Copyright © 2005 Pearson Education, Inc.

Application: Walking Speed

In a study by psychologists Bornstein and Bornstein, it was found that the average walking speed w, in feet per second, of a person living in a city of population P, in thousands, is given by the function

w(P) = 0.37 ln P + 0.05.

Page 33: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-33Copyright © 2005 Pearson Education, Inc.

Application: Walking Speed continued

The population of Philadelphia, Pennsylvania, is 1,517,600. Find the average walking speed of people living in Philadelphia.

Since 1,517,600 = 1517.6 thousand, we substitute 1517.6 for P, since P is in thousands: w(1517.6) = 0.37 ln 1517.6 + 0.05

2.8 ft/sec.The average walking speed of people living in Philadelphia is about 2.8 ft/sec.

Page 34: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Properties of Logarithmic Functions

· Convert from logarithms of products, powers, and quotients to expressions in terms of individual

logarithms, and conversely.· Simplify expressions of the type loga ax and .

loga xa

Page 35: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-35Copyright © 2005 Pearson Education, Inc.

Logarithms of Products

The Product Rule

For any positive numbers M and N and any logarithmic base a,

loga MN = loga M + loga N.

(The logarithm of a product is the sum of the logarithms of the factors.)

Page 36: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-36Copyright © 2005 Pearson Education, Inc.

Example

Express as a single logarithm: .

Solution:

23 3log logx w

2 23 3 3log log log ( )x w x w

Page 37: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-37Copyright © 2005 Pearson Education, Inc.

Logarithms of Powers

The Power Rule

For any positive number M, any logarithmic base a, and any real number p,

loga Mp = p loga M.

(The logarithm of a power of M is the exponent times the logarithm of M.)

Page 38: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-38Copyright © 2005 Pearson Education, Inc.

Examples

Express as a product. Express as a product.

3log 7a

3log 7 3log 7a a

5log 11a

1/ 55log 11 log 11

1log 11

5

a a

a

Page 39: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-39Copyright © 2005 Pearson Education, Inc.

Logarithms of Quotients

The Quotient Rule

For any positive numbers M and N, and any logarithmic base a,

.

(The logarithm of a quotient is the logarithm of the numerator minus the logarithm of the denominator.)

log log loga a a

MM N

N

Page 40: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-40Copyright © 2005 Pearson Education, Inc.

Examples

Express as a difference of logarithms.

Express as a single logarithm.

10log

log 10 log

a

a a

bb 125

log log 525w w

log 125 log 25w w

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Slide 4-41Copyright © 2005 Pearson Education, Inc.

Applying the Properties

Express in terms of sums and differences of logarithms.

3 43 4 2

2

3 4 2

log log ( ) log

log log log

3log 4log 2log

a a a

a a a

a a a

w yw y z

z

w y z

w y z

3 4

2loga

w y

z

Page 42: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-42Copyright © 2005 Pearson Education, Inc.

Example

Express as a single logarithm.

16log 2log log

3b b bx y z

6 2 1/ 3

61/ 3

2

6 1/ 3 6 3

2 2

16log 2log log log log log

3

log log

log , or log

b b b b b b

b b

b b

x y z x y z

xz

y

x z x z

y y

Page 43: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-43Copyright © 2005 Pearson Education, Inc.

Final Properties

The Logarithm of a Base to a PowerFor any base a and any real number x,

loga ax = x.(The logarithm, base a, of a to a power is the power.)

A Base to a Logarithmic PowerFor any base a and any positive real number x,

(The number a raised to the power loga x is x.)

log .a xa x

Page 44: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-44Copyright © 2005 Pearson Education, Inc.

Examples

Simplify.

a) loga a6

b) ln e8

Solution:

a) loga a6 = 6

b) ln e8 = 8

Simplify.

a)

b)

Solution:

a)

b)

7log7 w

7log7 w w

ln 8e

log 8ln 8 8ee e

Page 45: Slide 4-1 Copyright © 2005 Pearson Education, Inc

4.5 Solving Exponential and Logarithmic Equations

· Solve exponential and logarithmic equations.

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Solving Exponential Equations

Equations with variables in the exponents, such as

3x = 40 and 53x = 25, are called exponential equations.

Base-Exponent Property

For any a > 0, a 1,

ax = ay x = y.

Page 47: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-47Copyright © 2005 Pearson Education, Inc.

Example

Solve: .

Write each side with the same base.

Since the bases are the same number, 5, we can use the base-exponent property and set the exponents equal:

Check: 52x 3 = 125

52(3) 3 ? 125

53 ? 125

125 = 125 True

The solution is 3.

2 35 125x

2 3 35 5x

2 3 3

2 6

3

x

x

x

Page 48: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-48Copyright © 2005 Pearson Education, Inc.

Another Property

Property of Logarithmic Equality

For any M > 0, N > 0, a > 0, and a 1,

loga M = loga N M = N.

Solve: 2x = 50

log 2 log50

log 2 log50

log50

log 2

x

x

x

This is an exact answer. We cannot simplify further, but we can approximate using a calculator.

x 5.6439

We can check by finding 25.6439 50.

Page 49: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-49Copyright © 2005 Pearson Education, Inc.

Example

Solve: e0.25w = 12

The solution is about 9.94.

0.25

0.25

12

ln ln12

0.25 ln12

ln12

0.259.9396

w

w

e

e

w

w

w

Page 50: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-50Copyright © 2005 Pearson Education, Inc.

Solving Logarithmic Equations

Equations containing variables in logarithmic expressions, such as log2 x = 16 and log x + log (x + 4) = 1, are called logarithmic equations.

To solve logarithmic equations algebraically, we first try to obtain a single logarithmic expression on one side and then write an equivalent exponential equation.

Page 51: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-51Copyright © 2005 Pearson Education, Inc.

Example

Solve: log4 x = 3 Check:

log4 x = 3

The solution is

4

3

3

log 3

4

1

41

64

x

x

x

x

14 64

34

log ? 3

log 4 ? 3

3 3

1.

64

Page 52: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-52Copyright © 2005 Pearson Education, Inc.

Example

Solve: Check: For x = 3:

For x = 3:

2 2

2

3

2

2

log ( 1) log ( 1) 3

log ( 1)( 1) 3

( 1)( 1) 2

1 8

9

9

3

x x

x x

x x

x

x

x

x

2 2log ( 1) log ( 1) 3x x 2 2

2 2

2

2

log ( 1) log ( 1)?3

log 4 log 2?3

log (4 2)?3

log 8

3

?3

3 3

3

2 2

2 2

log ( 1) log ( 1)? 3

log ( 2) log ( 4)? 3

3 3

Negative numbers do not have real-number logarithms. The solution is 3.

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Slide 4-53Copyright © 2005 Pearson Education, Inc.

Example

Solve: ln ln( 4) ln3x x ln ln( 4) ln3

ln ln34

34

4 3( 4)4

3 12

12 2

6

x x

x

xx

xx

x xx

x x

x

x

The value 6 checks and is the solution.

Page 54: Slide 4-1 Copyright © 2005 Pearson Education, Inc

4.6 Applications and Models:

Growth and Decay

· Solve applied problems involving exponential growth and decay.

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Slide 4-55Copyright © 2005 Pearson Education, Inc.

Population Growth

The function P(t) = P0ekt, k > 0 can model many kinds of population growths.

In this function:

P0 = population at time 0,

P = population after time,

t = amount of time,

k = exponential growth rate.

The growth rate unit must be the same as the time unit.

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Slide 4-56Copyright © 2005 Pearson Education, Inc.

Example

Population Growth of the United States. In 1990 the population in the United States was about 249 million and the exponential growth rate was 8% per decade. (Source: U.S. Census Bureau) Find the exponential growth function. What will the population be in 2020? After how long will the population be double what it

was in 1990?

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Slide 4-57Copyright © 2005 Pearson Education, Inc.

Solution

At t = 0 (1990), the population was about 249 million. We substitute 249 for P0 and 0.08 for k to obtain the exponential growth function.

P(t) = 249e0.08t

In 2020, 3 decades later, t = 3. To find the population in 2020 we substitute 3 for t:

P(3) = 249e0.08(3) = 249e0.24 317.

The population will be approximately 317 million in 2020.

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Slide 4-58Copyright © 2005 Pearson Education, Inc.

Solution continued

We are looking for the doubling time T.

498 = 249e0.08T

2 = e0.08T

ln 2 = ln e0.08T (Taking the natural logarithm on both sides)

ln 2 = 0.08T (ln ex = x)

= T

8.7 T

The population of the U.S. will double in about 8.7 decades or 87 years. This will be approximately in 2077.

ln 2

0.08

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Slide 4-59Copyright © 2005 Pearson Education, Inc.

Interest Compound Continuously

The function P(t) = P0ekt can be used to calculate interest that is compounded continuously.

In this function:

P0 = amount of money invested,

P = balance of the account,

t = years,

k = interest rate compounded continuously.

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Example

Suppose that $2000 is deposited into an IRA at an interest rate k, and grows to $5889.36 after 12 years.

What is the interest rate? Find the exponential growth function. What will the balance be after the first 5 years? How long did it take the $2000 to double?

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Solution

At t = 0, P(0) = P0 = $2000. Thus the exponential growth function is P(t) = 2000ekt. We know that P(12) = $5889.36. We then substitute and

solve for k: $5889.36 = 2000e12k

The interest rate is about 9%.

12

12

5889.36

20005889.36

ln ln2000

5889.36ln 12

20005889.36

ln200012

0.09

k

k

e

e

k

k

k

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Slide 4-62Copyright © 2005 Pearson Education, Inc.

Solution continued

The exponential growth function is

P(t) = 2000e0.09t.

The balance after 5 years is

P(5) = 2000e0.09(5)

= 2000e0.45

$3136.62

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Slide 4-63Copyright © 2005 Pearson Education, Inc.

Solution continued

To find the doubling time T, set P(T) = 2 P0 = $4000 and solve for T.

4000 = 2000e0.09T

2 = e0.09T

ln 2 = ln e0.09T

ln 2 = 0.09T

= T

7.7 T The original investment of $2000 doubled in about 7.7 years.

ln 2

0.09

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Slide 4-64Copyright © 2005 Pearson Education, Inc.

Growth Rate and Doubling Time

The growth rate k and the doubling time T are related by

kT = ln 2

or

or

* The relationship between k and T does not depend on P0.

ln 2k

T

ln 2T

k

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Slide 4-65Copyright © 2005 Pearson Education, Inc.

Example

A certain town’s population is doubling every 37.4 years. What is the exponential growth rate?

Solution:

ln 2 ln 21.9%

37.4k

T

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Slide 4-66Copyright © 2005 Pearson Education, Inc.

Models of Limited Growth

In previous examples, we have modeled population growth. However, in some populations, there can be factors that prevent a population from exceeding some limiting value.

One model of such growth is

which is called a logistic function. This function increases toward a limiting value a as t approaches infinity. Thus, y = a is the horizontal asymptote of the graph.

( )1 kt

aP t

be

Page 67: Slide 4-1 Copyright © 2005 Pearson Education, Inc

Slide 4-67Copyright © 2005 Pearson Education, Inc.

Exponential Decay

Decay, or decline, of a population is represented by the function P(t) = P0ekt, k > 0.

In this function:

P0 = initial amount of the substance, P = amount of the substance left after time, t = time, k = decay rate.

The half-life is the amount of time it takes for half of an amount of substance to decay.

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Graphs

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Example

Carbon Dating. The radioactive element carbon-14 has a half-life of 5750 years. If a piece of charcoal that had lost 7.3% of its original amount of carbon, was discovered from an ancient campsite, how could the age of the charcoal be determined?

Solution: We know (from Example 5 in our book), that the function for carbon dating is

P(t) = P0e-0.00012t.

If the charcoal has lost 7.3% of its carbon-14 from its initial amount P0, then 92.7%P0 is the amount present.

Page 70: Slide 4-1 Copyright © 2005 Pearson Education, Inc

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Example continued

To find the age of the charcoal, we solve the equation for t :

The charcoal was about 632 years old.

0.000120 0

0.00012

0.00012

92.7%

0.927

ln 0.927 ln

ln 0.927 0.00012

ln 0.927

0.00012632

t

t

t

P P e

e

e

t

t

t