Exponential Functions Define an exponential function. Graph exponential functions. Use transformations on exponential functions. Define simple interest

Exponential FunctionsDefine an exponential function.Graph exponential functions.Use transformations on exponential functions.Define simple interest.Develop a compound interest formula. Understand the number e.

SECTION 4.1

1

2

3

4

5

6

EXPONENTIAL FUNCTION

A function f of the form

is called an exponential function with base a. Its domain is (–∞, ∞).

f x ax , a 0 and a 1,

EXAMPLE 1 Evaluating Exponential Functions

a. Let f x 3x 2. Find f 4 .

b. Let g x 210x. Find g 2 .

c. Let h x 1

9

x

. Find h 3

2

.

d. Let F(x) = 4x. Find F(3.2).

EXAMPLE 1 Evaluating Exponential Functions

Solution

4 2 2a. 94 3 3f

22

1 1b. 2 10 2 2 0.02

10 1002g

33

1 2

3

22

1c. 9 9

3

227

9h

d. F(3.2) = 43.2 ≈ 84.44850629

RULES OF EXPONENTS

Let a, b, x, and y be real numbers with a > 0 and b > 0. Then

,x y x ya a a

,x

x yy

aa

a

,x x xab a b

,yx xya a

0 1,a

1 1.

xx

xa

a a

EXAMPLE 2Graphing an Exponential Function with Base a > 1 – Exponential Growth

Graph the exponential function

Solution

Make a table of values.

f x 3x.

Plot the points and draw a smooth curve.

EXAMPLE 2Graphing an Exponential Function with

Base a > 1

Solution continued

This graph is typical for exponential functions when a > 1.

EXAMPLE 3Graphing an Exponential Function with Base 0 < a < 1 – Exponential Decay

Sketch the graph of

Solution


1.

2

x

y

Plot the points and draw a smooth curve.

EXAMPLE 3Graphing an Exponential Function with

Base 0 < a < 1

Solution continued

As x increases in the positive direction, y decreases towards 0.

PROPERTIES OF EXPONENTIAL FUNCTIONS

Let f (x) = ax, a > 0, a ≠ 1.

1. The domain of f (x) = ax is (–∞, ∞).

2. The range of f (x) = ax is (0, ∞); the entire graph lies above the x-axis.

3. For a > 1, Exponential Growth (i) f is an increasing function, so the graph rises to

the right.

(ii) as x → ∞, y → ∞.

(iii) as x → –∞, y → 0.

4. For 0 < a < 1, - Exponential Decay (i) f is a decreasing function, so the graph falls to the

right.(ii) as x → – ∞, y → ∞.

(iii) as x → ∞, y → 0.

5. The graph of f (x) = ax has no x-intercepts, so it never crosses the x-axis. No value of x will cause f (x) = ax to equal 0.

6. The graph of is a smooth and continuous curve, and it passes through the points

7. The x-axis is a horizontal asymptote for every exponential function of the form f (x) = ax.

TRANSFORMATIONS ON EXPONENTIAL

FUNCTION f (x) = ax Transformation Equation Effect on Equation

HorizontalShift

y = ax+b

Shift the graph of y = ax, |b| units(i) left if b > 0.(ii) right if b < 0.

VerticalShift

y = ax + b

Shift the graph of y = ax, |b| units(i) up if b > 0.(ii) down if b < 0.



Stretching or Compressing(Vertically)

y = cax

Multiply the y coordinates by c. The graph of y = ax is vertically(i) stretched if c > 1.(ii) compressed if 0 < c < 1.



Reflection y = –ax Reflect the graph of y = ax in the x-axis.

Reflect the graph of y = ax in the y-axis.

y = a–x

EXAMPLE 6 Sketching Graphs

Use transformations to sketch the graph of each function.

3 4xf x a.

State the domain and range of each function and the horizontal asymptote of its graph.

1 3xf x b.

3xf x c. 3 2xf x d.


Solution

Domain: (–∞, ∞)

Range: (–4, ∞)

Horizontal Asymptote: y = –4

3 4xf x a.


Solution continued


Range: (0, ∞)

Horizontal Asymptote: y = 0

1 3xf x b.


Solution continued


Range: (–∞, 0)


3xf x c.


Solution continued


Range: (–∞, 2)


3 2xf x d.

General Exponential Growth/Decay Model

Original amount

Rate of decay (r < 0),Growth (r > 0)

Number of time periods

Amount after t time periods

Compound interest is the interest paid on both the principal and the accrued (previously earned) interest. It is an application of exponential growth.

Interest that is compounded annually is paid once a year. For interest compounded annually, the amount A in the account after t years is given by

COMPOUND INTEREST – Growth

Amount after t time periods

Original amount

Rate of decay (r < 0),Growth (r > 0)

Number of time periods

EXAMPLE 2 Calculating Compound Interest

Juanita deposits $8000 in a bank at the interest rate of 6% compounded annually for five years.

a. How much money will she have in her account after five years?

b. How much interest will she receive?

EXAMPLE 2 Calculating Compound Interest

Solution

a. Here P = $8000, r = 0.06, and t = 5.

b. Interest = A P = $10,705.80 $8000 = $2705.80.

COMPOUND INTEREST FORMULA

A = amount after t yearsP = principalr = annual interest rate (expressed as a decimal)n = number of times interest is compounded

each yeart = number of years

1nt

rA P n

EXAMPLE 3Using Different Compounding Periods to Compare Future Values

If $100 is deposited in a bank that pays 5% annual interest, find the future value A after one year if the interest is compounded

(i) annually.(ii) semiannually.(iii) quarterly.(iv) monthly.(v) daily.


(i) Annual Compounding:

1

1

1 $0.05 1 00 05 00 .

ntr

A Pn

A

Solution

In the following computations, P = 100, r = 0.05 and t = 1. Only n, the number of times interest is compounded each year, changes. Since t = 1, nt = n(1) = n.


(iii) Quarterly Compounding:4

4

10

14

10

$105.0.05

40 9

rA P

A

(ii) Semiannual Compounding:2

2

10

1

1 $105.060

2

.050

rA P

n

A


(iv) Monthly Compounding:1

12

2

1012

112

1 $10505

..0

120

rA P

A

(v) Daily Compounding:3

365

65

100.

1365

1 $1365

. 305

05 10

rA P

A

EXAMPLE 8 Bacterial Growth

A technician to the French microbiologist Louis Pasteur noticed that a certain culture of bacteria in milk doubles every hour. If the bacteria count B(t) is modeled by the equation

B t 20002t ,

a. the initial number of bacteria,b. the number of bacteria after 10 hours; andc. the time when the number of bacteria will be

32,000.

with t in hours, find

EXAMPLE 8 Bacterial Growth

00 2000 2 200 00 1 2000B B

a. Initial size

10b. 2000 2 2,10 048,000B

32,000 2000 2

16 2

t

t

c. Find t when B(t) = 32,000

24 2t

4 t

After 4 hours, the number of bacteria will be 32,000.

Solution

THE VALUE OF e

The value of e to 15 places is e = 2.718281828459045.

gets closer and closer to a fixed number. This irrational number is denoted by e and is sometimes called the Euler number.

As h gets larger and larger,

11

h

h

CONTINUOUS COMPOUND FORMULA

A = amount after t yearsP = principalr = annual rate (expressed as a decimal)t = number of years

ertA P

EXAMPLE 4 Calculating Continuous Compound Interest

Find the amount when a principal of $8300 is invested at a 7.5% annual rate of interest compounded continuously for eight years and three months.Solution

P = $8300 and r = 0.075. Convert eight years and three months to 8.25 years.

0.07 85 .25

$15,409.

$8300

83

rtA Pe

A e

EXAMPLE 5 Calculating the Amount of Repaying a Loan

How much money did the government owe DeHaven’s descendants for 213 years on a $450,000 loan at the interest rate of 6%?

Solution

a. With simple interest,

0.0$450,00

1

0 1

$6.201 million.

6 213

A P Prt P rt

A


Solution continued

b. With interest compounded yearly, 1

11

2 31 1

$1.105 10

$450,00

$110.500 millio

0.

n.

0 06t

A P r

A

c. With interest compounded quarterly, 4 4 21

11

30.0

1 14 4

$1.45305 10

$145.30

$450,

5 billi

0

o .

600

n

tr

A P

A


Solution continued

d. With interest compounded continuously,

Notice the dramatic difference between quarterly and continuous compounding and the dramatic difference between simple interest and compound interest.

2130.0

11

6

$1.5977 1

$450,00

0

$159.77 billion.

0rtA Pe e

A

THE NATURAL EXPONENTIAL FUNCTION

with base e is so prevalent in the sciences that it is often referred to as the exponential function or the natural exponential function.

xf x e

The exponential function

EXAMPLE 6 Sketching a Graph

Use transformations to sketch the graph of

Solution

Start with the graph of y = ex.



Solution coninued

Shift the graph of y = ex one unit right.



Solution continued

Shift the graph of y = ex – 1 two units up.

MODEL FOR EXPONENTIALGROWTH OR DECAY

0ktA t A e

A(t) = amount at time t A0 = A(0), the initial amount k = relative rate of growth (k > 0) or decay

(k < 0) t = time

EXAMPLE 7 Modeling Exponential Growth and Decay

In the year 2000, the human population of the world was approximately 6 billion and the annual rate of growth was about 2.1%. Using the model on the previous slide, estimate the population of the world in the following years.

a. 2030b. 1990

EXAMPLE 7

300.0213

11.2 5 3

6

66

0

6

A e

a. The year 2000 corresponds to t = 0. So A0 = 6 (billion), k = 0.021, and 2030 corresponds to t = 30.

Solution

The model predicts that if the rate of growth is 2.1% per year, over 11.26 billion people will be in the world in 2030.

Modeling Exponential Growth and Decay

EXAMPLE 7

0.021 10

4.863505

6

5

10A e

b. The year 1990 corresponds to t = 10.

Solution

The model predicts that the world had over 4.86 billion people in 1990. (The actual population in 1990 was 5.28 billion.)

Modeling Exponential Growth and Decay

Logarithmic Functions

Define logarithmic functions.Inverse FunctionsEvaluate logarithms.Rules of LogarithmsFind the domains of logarithmic functions.Graph logarithmic functions.Use logarithms to evaluate exponential equations.

SECTION 4.3

1

2

3

4

5

6

7

DEFINITION OF THELOGARITHMIC FUNCTION

For x > 0, a > 0, and a ≠ 1,

y loga x if and only if x ay .

The function f (x) = loga x, is called the logarithmic function with base a.

The logarithmic function is the inverse function of the exponential function.

Inverse Functions

Certain pairs of one-to-one functions “undo” one another. For example, if

5( ) 8 5 and ( ) ,

8x

x x x f g

then

855)10(8)10( f 108

5)85()85(

g

Inverse Functions

Starting with 10, we “applied” function and then “applied” function g to the result, which returned the number 10.

Inverse Functions

As further examples, check that

( ) and (3 29 2 ,3)9 f g

( ) and ( 3 5)3 5 ,55 f g

( ) and 3 38 8

2 ,2

g g

Inverse Functions

( ( )) and ( (2 2 2 ) .2) f g g f

In particular, for this pair of functions,

In fact, for any value of x,

( ( )) and ( ( )) ,x x x x f g g f

or ( )( ) and ( )( ) .x x x x f g g f

Because of this property, g is called the inverse of .

Inverse Function

Let be a one-to-one function. Then g is the inverse function of if

( )( ) x xf g for every x in the domain of g,

( )( )x xg f

andfor every x in the domain of .

EXAMPLE 1Converting from Exponential to Logarithmic Form

a. 43 64

Write each exponential equation in logarithmic form.

b. 1

2

4

1

16c. a 2 7

Solution

43a. 4 64 3log 64

4

1 2

1 1 1b. log

2 16 14

6

2 2c. 7 log 7aa

EXAMPLE 2Converting from Logarithmic Form to Exponential Form

a. log3 2435

Write each logarithmic equation in exponential form.

b. log2 5 x c. loga N x

Solution

35a. log 243 5 243 3

2b. log 5 5 2xx

c. logaxN x N a

EXAMPLE 3 Evaluating Logarithms

a. log5 25

Find the value of each of the following logarithms.

b. log2 16 c. log1 3 9

d. log7 7 e. log6 1 f. log4

1

2

Solution2

5a. log 25 25 5 or 5 5 2yyy y

42b. log 16 16 2 or 2 2 4yyy y

EXAMPLE 3 Evaluating Logarithms

Solution continued

17d. log 7 7 7 or 7 7 1y yy y

06e. log 1 1 6 or 6 6 0y yy y

1 24

1 1 1f. log 4 or 2 2

2 2 2y y yy

21 3

1c. log 9 9 or 3 3 2

3y

y

y y

EXAMPLE 4 Using the Definition of Logarithm

a. log5 x 3

Solve each equation.

b. log3

1

27y

c. logz 1000 3 22d. log 6 10 1x x

Solution

5

3

3

3a. log

5

1 1

5 125

x

x

x


Solution continued

3

3

1b. log

271

327

3 3

3

y

y

y

y

3 3

3

c. log 1000

1000

1

0

3

0

1

z

z

z

z


Solution continued

22

2

2

1

d. log 6 10

6 10 2 2

6 8 0

2

1

4 0

x x

x x

x x

x x

x 2 0 or x 4 0

x 2 or x 4

Rules of Logarithms with Base aIf M, N, and a are positive real numbers with a ≠ 1, and x is

any real number, then

1. loga(a) = 1 2. loga(1) = 0

3. loga(ax) = x 4.

5. loga(MN) = loga(M) + loga(N)

6. loga(M/N) = loga(M) – loga(N)

7. loga(Mx) = x · loga(M) 8. loga(1/N) = – loga(N)

Na Na )(log

Rules of Logarithms

These relationships are used to solve exponential or logarithmic equations

COMMON LOGARITHMS

1. log 10 = 1

2. log 1 = 0

3. log 10x = x4. 10log x x

The logarithm with base 10 is called the common logarithm and is denoted by omitting the base: log x = log10 x. Thus,

y = log x if and only if x = 10 y.

Applying the basic properties of logarithms

NATURAL LOGARITHMS

1. ln e = 1

2. ln 1 = 0

3. log ex = x4. eln x x

The logarithm with base e is called the natural logarithm and is denoted by ln x. That is, ln x = loge x. Thus,

y = ln x if and only if x = e y.

Applying the basic properties of logarithms

DOMAIN OF LOGARITHMIC FUNCTION

Domain of y = loga x is (0, ∞)Range of y = loga x is (–∞, ∞)

Logarithms of 0 and negative numbers are not defined.

EXAMPLE 5 Finding the Domain

3log 2f x x Find the domain of

2 0

2

x

x

Solution

Domain of a logarithmic function must be positive, that is,

The domain of f is (–∞, 2).


Sketch the graph of y = log3 x.Solution by plotting points (Method 1)



Solution continued

Plot the ordered pairs and connect with a smooth curve to obtain the graph of y = log3 x.

66


Solution by using the inverse function (Method 2)

Graph y = f (x) = 3x.

Reflect the graph of y = 3x in the line y = x to obtain the graph of y = f –1(x) = log3 x.

GRAPHS OF LOGARITHMIC FUNCTIONS

PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Exponential Function f (x) = ax

Logarithmic Function f (x) = loga x

Domain (0, ∞) Range (–∞, ∞)

1. Domain (–∞, ∞) Range (0, ∞)

x-intercept is 1 No y-intercept

2. y-intercept is 1 No x-intercept

3. x-axis (y = 0) is the horizontal asymptote

y-axis (x = 0) is the vertical asymptote




The graph is a continuous smoothcurve that passes through the points

(1, 0), and

(a, 1).

4. The graph is a continuous smooth curve that passes through the points

(0, 1), and

(1, a).

11, ,

a

1, 1 ,

a




Is one-to-one, that is, logau = logav if and only if u = v.

5. Is one-to-one , that is, au = av if and only if u = v. Increasing if a > 1

Decreasing if 0 < a < 16. Increasing if a > 1 Decreasing if 0 < a < 1

EXAMPLE 7 Using Transformations

Start with the graph of f (x) = log3 x and use transformations to sketch the graph of each function.

a. f x log3 x 2

c. f x log3 x

b. f x log3 x 1

d. f x log3 x

State the domain and range and the vertical asymptote for the graph of each function.


Solution

Shift up 2Domain (0, ∞)Range (–∞, ∞)Vertical asymptote x = 0

3a. l 2ogf x x


Solution continued

Shift right 1Domain (1, ∞)Range (–∞, ∞)Vertical asymptote x = 1

3b. l 1ogf x x


Solution continued

Reflect graph of y = log3 x in the x-axis Domain (0, ∞)Range (–∞, ∞)Vertical asymptote x = 0

3c. logf x x


Solution continued

Reflect graph of y = log3 x in the y-axis Domain (∞, 0)Range (–∞, ∞)Vertical asymptote x = 0

3d. logf x x

EXAMPLE 8 Using Transformations to Sketch a Graph

Sketch the graph of y 2 log x 2 .Solution

Start with the graph of f (x) = log x.

Step 1: Replacing x with x – 2 shifts thegraph two units right.

EXAMPLE 8 Using Transformations to Sketch a Graph

Solution continued

Step 2: Multiplyingby 1 reflects the graph

Step 3: Adding 2 shifts the graph

two units up.in the x-axis.

Rules of Logarithms with Base aIf M, N, and a are positive real numbers with a ≠ 1, and x is

any real number, then

1. loga(a) = 1 2. loga(1) = 0

3. loga(ax) = x 4.

5. loga(MN) = loga(M) + loga(N)

6. loga(M/N) = loga(M) – loga(N)

7. loga(Mx) = x · loga(M) 8. loga(1/N) = – loga(N)

Na Na )(log

Rules of Logarithms

EXAMPLE 1Using Rules of Logarithms to Evaluate Expressions

5a. log yz

Given that log 5 z = 3 and log 5 y = 2, evaluate each expression.

75b. log 125y

5c. logz

y 1/30 55d. log z y

Solution 55 5a. log

3

l loog

2

5

gyz y z


7 75 5 5

5 53

b. log 125 log 125 log

lo logg 5 7

3 1727

y

y y

1/2

5 555 log

2

1c. log lo logg

2

1

2 23

1

z z

yz y

y

Solution continued


Solution continued

5

1/30 5 1/30 55 5 5

5

d. log log log

15

301

530

log

3

0.1 10

10.

2

1

log

z y z y

z y

EXAMPLE 2 Writing Expressions In Expanded Form

32

2 4

1a. log

2 1

x x

x

Write each expression in expanded form.

3 2 5b. ln x y z

Solution

323 42

2 2 24

3 422 2 2

2 2 2

1a. log log 1 log 2 1

2 1

log log 1 log 2 1

2log 3log 1 4log 2 1

x xx x x

x

x x x

x x x

EXAMPLE 2 Writing Expressions In Expanded Form

1/23 2 5 3 2 5

3 2 5

3 2 5

b. ln ln

1ln

21

ln ln ln21

3ln 2ln 5ln23 5

ln ln ln2 2

x y z x y z

x y z

x y z

x y z

x y z

Solution continued

EXAMPLE 3 Writing Expressions in Condensed Form

a. log3 log 4x y

Write each expression in condensed form.

21b. 2ln ln 1

2x x

2 2 2c. 2log 5 log 9 log 75

21d. ln ln 1 ln 1

3x x x


3a. log3 log 4 log

4

xx y

y

2 2

2 2

1/22b. 1

ln ln 1 ln ln 1

ln 1

22 x x x x

x x

Solution


2 2 2

2 2 2

2 2

2

2

2

c. 2log 5 log 9 log 75

log log log 75

log log 75

25 9log

75lo

5

g

9

25 9

3

Solution continued


Solution continued

2

2

2

32

1d. ln 1

31

ln 13

11ln

3 1

ln ln 1

ln

1n

1

l1

x

x

x x

x

x x

x

x x

x x

CHANGE-OF-BASE FORMULA

Let a, b, and x be positive real numbers with a ≠ 1 and b ≠ 1. Then logb x can be converted to a different base as follows:

log log lnlog

log log ln

(ba base 1se ) ( ) ( )bas0 e

ab

a

x x x

e

xb b

a

b

EXAMPLE 4Using a Change of Base to Compute Logarithms

Compute log513 by changing to a. common logarithms and b. natural logarithms.

5

13lnb.

513 log

ln1.59369

5

13loga. log

log

1.5

13

69

5

93

Solution

EXAMPLE 9 Evaluating the Natural Logarithm

Evaluate each expression.

a. ln e4 b. ln1

e2.5 c. ln 3

Solution

4a. ln 4e

2.2.5

51b. ln l .5n 2e

e

Use a calculator.c. ln 31.0986123

EXAMPLE 10 Doubling Your Money

a. How long will it take to double your money if it earns 6.5% compounded continuously?

b. At what rate of return, compounded continuously, would your money double in 5 years?

Solutiona. If P is the original

amount invested, A = 2P.

It will take 11 years to double your money.

EXAMPLE 10 Doubling Your Money

Solution continued

b. Your investment will double in 5 years at the rate of 13.86%.

Solving Exponential Or Logarithmic EquationsTo solve an exponential or logarithmic equation, change the given equation into one of the following forms, where a and b are real numbers, a > 0 and a ≠ 1, and follow the guidelines.1.ax = b Solve by taking logarithms on both sides.2. Loga x = b Solve by changing to exponential form ab = x.

SOLVING AN EXPONENTIAL EQUATION

Solve 7x = 12. Give the solution to the nearest thousandth.

Solution

While any appropriate base b can be used, the best practical base is base 10 or base e. We choose base e (natural) logarithms here.


Solve 7x = 12. Give the solution to the nearest thousandth.

Solution 7 12x

I 7 12n Inx Property of logarithms

In 7 In 12x Power of logarithms

In12In 7

x Divide by In 7.

1.277x Use a calculator.

The solution set is {1.277}.


Solve 32x – 1 = .4x+2 . Give the solution to the nearest thousandth.

Solution 2 1 23 .4x x

2 1 2In In 3 .4x x Take natural logarithms on both sides.

(2 1) In 3 ( 2) In .4x x Property power

2 In 3 In 3 In .4 2 In .4x x Distributive property


Solve 32x – 1 =.4x+2 . Give the solution to the nearest thousandth.

Solution

2 In 3 In .4 2 In .4 In 3x x Write the terms with x on one side

(2 In 3 In .4) 2 In .4 In 3x Factor out x.

2 In .4 32 In 3 .4

x

Divide by 2 In 3 – In .4.

2

2

In .4 In 3In 3 In .4

x

Power property


Solve 32x – 1 =.4x+2 . Give the solution to the nearest thousandth.

Solution In .16 In 3

In 9 In .4x

Apply the exponents.

In .489

In .4

x Product property; Quotient property

.236x This is approximate.

This is exact.

The solution set is { –.236}.

SOLVING BASE e EXPONENTIAL EQUATIONS

Solve the equation. Give solutions to the nearest thousandth.

Solution

a.2

200x e

2

200x e

2

In In 200x e Take natural logarithms on both sides.

2 In 200x In = x22xe


Solution

a.2

200x e

Square root propertyIn 200x

Remember both roots.

2.302x Use a calculator.

The solution set is { 2.302}.



Solution

b.

Take natural logarithms on both sides.

2 1 3x e e m n m na a a

2In In 3x e

2 In In 3x e Power property


2 3x e Divide by e; .m

m nn

aa

a

eee xx 3412

eee xx 3412


Solution

b.

2 In 3x In e = 1

1In 3

2x Multiply by – ½

.549x

The solution set is {– .549}.


eee xx 3412

SOLVING A LOGARITHMIC EQUATION

Solve log(x + 6) – log(x + 2) = log x.

Solution

log( 6) log( 2) logx xx

lo6

og2

glxx

x

Quotient property

62

xx

x

Property of logarithms

6 ( 2)x x x


Solve log(x + 6) – log(x + 2) = log x.

Solution Distributive property

26 2x x x

Standard form2 6 0x x

( 3)( 2) 0x x Factor.

3 or 2x x Zero-factor property

The proposed negative solution (x = – 3) is not in the domain of the log x in the original equation, so the only valid solution is the positive number 2, giving the solution set {2}.


Solve log(3x + 2) + log(x – 1 ) = 1. Give the exact value(s) of the solution(s).

Solution log(3 2) l 1 1og( )x x

log(3 2) log( 1) log10x x Substitute.

log[(3 2)( 1)] log10x x Product property

(3 2)( 1) 10x x Property of logarithms

SOLVING A LOGARITMIC EQUATION


Solution 23 2 10x x Multiply.

23 12 0x x Subtract 10.

1 1 1446

x Quadratic formula

SOLVING A LOGARITMIC EQUATION


Solution

1 1456

The number is negative, so x – 1 is negative. Therefore, log(x – 1) is not defined and this proposed solution must be discarded.

Since > 1, both 3x + 2 and x – 1 are positive and the solution set is

1 1456

1 145.

6

NEWTON’S LAW OF COOLING

Newton’s Law of Cooling states that

where T is the temperature of the object at time t, Ts is the surrounding temperature, and T0 is the value of T at t = 0.

T Ts T0 Ts e kt ,

EXAMPLE 11 McDonald’s Hot Coffee

The local McDonald’s franchise has discovered that when coffee is poured from acoffeemaker whose contents are 180ºF into a noninsulated pot, after 1 minute, the coffee cools to 165ºF if the room temperature is 72ºF. How long should the employees wait before pouring the coffee from this noninsulated pot into cups to deliver it to customers at 125ºF?


Use Newton’s Law of Cooling with T0 = 180 and Ts = 72 to obtain

Solution

We have T = 165 and t = 1.

1807

7 08

2

1

72

2

kt

kt

T e

T e

72 108

93

1 8

165

0

k

k

e

e

93ln

108

0.1495317

k

k


Substitute this value for k.

Solution continued

Solve for t when T = 125.

T 72 108e 0.1495317t

0.1495317

0.1495317

1 72 108

125 72

10853

ln 0.14953 7

5

8

2

110

t

t

e

e

t

1 53ln

0.1495317 108

4.76

t

t

The employee should wait about 5 minutes.

GROWTH AND DECAY MODEL

A is the quantity after time t.A0 is the initial (original) quantity (when t = 0).r is the growth or decay rate per period.t is the time elapsed from t = 0.

0rtA A e

EXAMPLE 12 Chemical Toxins in a Lake

A chemical spill deposits 60,000 cubic meters of soluble toxic waste into a large lake. If 20% of the waste is removed every year, how many years will it take to reduce the toxin to 1000 cubic meters?

Solution

In the equation A = A0ert, we need to find A0, r, and the time when A = 1000.


60,000 rtA e

1. Find A0. Initially (t = 0), we are given A0 = 60,000. So

Solution continued

2. Find r. When t = 1 year, the amount of toxin will be 80% of its initial value, or


Solution continued

2. continued So


Solution continued

3. Find t when A = 1000.

It will take approximately 18 years to reduce toxin to 1000 m3.

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