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3.1: Exponential and Logistic Functions.

I. Exponential functions and their graphs.A. Let a and b be real number constants. An exponential

function in x is a function that can be written in the form:

B. Notes:a is the initial value of f (at x=0)

( ) xf x ab= 0, 0, 1a b b≠ > ≠

or recursively, ( 1) ( )f x b f x+ = ⋅

f ( )b is the base.When 0<b<1, it is exponential decayWhen 1<b, it is exponential growth.Domain is all real numbersRange is all non-negative real numbers.

C. Example 1: which of the following are exponential functions?

Function Yes/no

1.

2.

3.

( ) 2xf x =

( )1( ) 2x

h x =

3( )g x x= noyes

yes

4.

5.

6.

7.

( )( ) 2

( )( ) 3 xf x = −

( ) 5x

f x =

( ) 1xf x =

( ) 3f x π=

yesnononoy

D. Example 2: evaluate each of the following for the function below:

1. (5)f

2. (0)f

3. ( 2)f −

1⎛ ⎞

53 3 3 3 3 3= = ⋅ ⋅ ⋅ ⋅

03 1= =2

2

1 133 9

−= = =

14. 3

f ⎛ ⎞⎜ ⎟⎝ ⎠

1333 3 1.442= = ≈

25. 3

f ⎛ ⎞−⎜ ⎟⎝ ⎠

23

2 3 23

1 13 0.480733

−= = = ≈

6. ( )f π 3 31.5443π= ≈

E. Example 3: Find the exponential equation for each of the functions below.

12

10

8

6

4

2

f(x)

-2,34( )

2,12( )

1,6( )

0,3( )-1,32( )

Values for two exponential functions

x f(x) g(x)

-2 3/4 64/3

-1 3/2 16

30

20

10

-2 -1 1 2

g(x)-2,

643( )

-1,16( )

1,9( )0,12( )

2,274( )

-2 -1 1 2

0 3 12

1 6 9

2 12 27/4

( ) 3 2

3( ) 124

x

x

f x

g x

= ⋅

⎛ ⎞= ⎜ ⎟⎝ ⎠

F. The Natural Base e.1. The function is an example of an exponential

growth function.2. Definition: The Natural Base e.

3 Theorem: Exponential Functions and the Base e

( ) xf x e=

1lim 1x

xe

x→∞

⎛ ⎞= +⎜ ⎟⎝ ⎠

2.718281828459...e ≈3. Theorem: Exponential Functions and the Base e.

a. Any exponential function can be written as:

For some real number k.

b. If a>0 and k>0, is an exponential growth function

c. If a>0 and k<0, is an exponential decay function

( ) xf x ab=

( ) kxf x ae=

( ) kxf x ae=

( ) kxf x ae=

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II. Logistic functions and their graphs.A. Let a, b, c, and k be positive constants, with b < 1. A logistic

growth function in x is a function that can be written in the form

Where the constant c is the limit to growth.

( ) or ( )1 1x kx

c cf x f xab ae−= =

+ +g

Note: if b > 1 or k < 0, these formulas yield logistic decay functions.

B. Example 5: Graphing logistic growth functions.

( )4a. ( )

1 2 0.3 xf x =+ ( ) ( )0

int :4 4 4(0)

1 2 1 31 2 0.3

y

f

−

= = =++

. . : 0, 4H A y y= =

4g x( ) = 0

3

2

1

-4 -2 2 4

h x( ) = 4

f x( ) = 4

1+2⋅0.3x

2

30b. ( )1 4 xg x

e−=+ ( ) ( )2 0

int :30 30(0) 6

1 4 11 4

y

fe−

−

= = =++

. . : 0, 30H A y y= =

30h x( ) = 30

20

10

-20 20

g x( ) = 30

1+4⋅e-2 ⋅x

III. Population modelsA. Example 6: The population in Texas in 1985 was approximately

16,370,000 and about 25,726,000 in the year 2000.1. Find an exponential model that fits to the data above.

2. Estimate the Texas population in 2010.

16,370,000a =

1525726000 1.030616370000

b = ≈

3. In what year will the population exceed 40,000,000? (solve graphically)

( ) 16,370,000 1.0306tP t = ⋅25(25) 16,370,000 1.0306

(25) 34,774,000

P

P people

≈ ⋅

≈

29.646t yrs≈during the year 2014

B. Example 7: Suppose that a long-lasting epidemic spreads through a town with a population of 100,000 according to the logistic function below:

Where N(t) is the number of infected individuals, and t is the number of days after the initial infection.

1. How much of the population will be infected after 5 days?

100,000( )1 5000 tN t

e−=+

2. After how many days will half of the townspeople be infected? (solve graphically)

(5) 2,883N people≈

8.5days

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3.2: EXPONENTIAL AND LOGISTICMODELING

I. Constant Percentage Rate and Exponential Functions.

A. Exponential Population Model:1. If a population P is changing at a constant percentage

rate r each year, then

where P0 is the initial population, r is the growth rate and is expressed as a decimal, and t is time in years.if i iti ( d 1 1) thi i ti l

( )0( ) 1 tP t P r= +

2. if r is positive (and 1 + r > 1), this is an exponential growth function

3. if r is negative (and 1 + r < 1), this is an exponential decay function

B. Example 1: finding growth and decay ratesTell whether the population model is an exponential growth

or decay function, and find the constant percentage rate of growth or decay:

a. ( ) 32,459 0.9845tP t = ⋅decay, since 1 1

0.0155 decay rate is 1.55%

rr

+ <= −

growth, since 1 1r+ >b. ( ) 890,234 1.0239tP t = ⋅ 0.0239

growth rate is 2.39%r =

C. Example 2: Find the exponential function with the initial value of 28, and decaying at a rate of 4.5%.

D. Example 3: Find the exponential function containing the points (3,375) and (6,3).

( ) ( )( ) 28 1 0.045 28 0.955x xf x = − =

3(3) 375f ab⎧ = =⎪ 3⎛ ⎞

6

(3) 375(6) 3

f abf ab

⎧⎪⎨

= =⎪⎩

33753

b−=

15

b =

31(3) 1255

f a⎛ ⎞= = ⎜ ⎟⎝ ⎠

15625a =

1( ) 156255

x

f x ⎛ ⎞= ⎜ ⎟⎝ ⎠

II. Exponential Growth and Decay ModelsA. Doubling Time

Where f(t) is the population/amount, a is the initial amount, t is the time, and d is the doubling time.

B. Half-Life

( )( ) 2 2t t

ddf t a a= ⋅ =

1 1( )2 2

t th

hf t a a⎛ ⎞⎛ ⎞= ⋅ = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Where f(t) is the population/amount, a is the initial amount, t is the time, and h is the half-life.

2 2⎜ ⎟⎝ ⎠ ⎝ ⎠

C. Example 3: Radioactive iodine is used in various medical diagnostic tests. Its half-life is 8 days.

1. Write a formula for f(t)—the amount of iodine—in terms of the number of days.

2. What proportion of the original radioactive iodine is present after 48 days?

81( )2

t

f t a⎛ ⎞= ⎜ ⎟⎝ ⎠

48 681 1 1(48)f a a a⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

3. If there is initially 12mL of iodine present, after how many days will there be less than 1 mL? (solve graphically)

(48)2 2 64

f a a a= ⋅ = ⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

28.68t days≈

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D. Example 4: suppose that a colony of bacteria exhibits exponential growth. The colony initially contains 5000 bacteria, and takes 4 hours to double.

1. Write a function for P(t)—the population of bacteria—as a function of time.

2. How many bacteria are there after 10 hours?

( )4( ) 5000 2t

P t =

( )104(10) 5000 2 28 284P ( ) 4(10) 5000 2 28,284P = ≈

3. How long will it take until the colony contains 160,000 bacteria?

4

Algebraically:

5000 2 160,000t

⋅ =

42 32t

=t

note:this only worked out( )P t

542 2t

=

54t

∴ =

20t hours=

( )because is a power of 2P ta

III. Using Regression to Model Population

A. Example 5: The world population is given in the table to the right.

1. Find an exponential model for the data (in years after 1900)

2. Is this a reasonable model?( )( ) 1300.7007 1.0149 tP t ≈

YearPopulation (millions)

1950 2,555

1960 3,039

1970 3,707

1980 4,456

1990 5 2833. Predict the population in 2010

2, 0.9804 1yes r ≈ ≈

(210) 6,629,000,000 peopleP ≈

1990 5,283

2000 6,080

2010 6,823*

2020 7,518*

2030 8,140*

*projections

B. Example 6: The world population is given in the table to the right.

1. Find a logistic model for this population data. Use years after 1900 for t.

2. Use your model to predict the population in the year 2050

YearPopulation (millions)

1950 2,555

1960 3,039

1970 3,707

1980 4,456

1990 5 283

0.029

10929.493( )1 14.364 tP t

e−=+

(150) 9175million peopleP ≈3. Use this model to predict the maximum

sustainable population for the world.

1990 5,283

2000 6,080

2010 6,823*

2020 7,518*

2030 8,140*

*projections

lim ( ) 10,929,493,000t

P t→∞

=

IV. Other Logistic ModelsA. Example 7: In a college with a population of 800, a rumor

spreads according to the logistic equation below:

1. How many people will have heard a rumor after 1 minute? After 15 minutes?

0.4

800( )1 132 tP t

e−=+ ⋅

9 603≈ ≈2. When will the rumor reach at least 750 individuals? (solve

graphically)

9, 603≈ ≈

18min

B. Example 8: Find the logistic model with the following characteristics:

1. Initial value of 162. Maximum capacity of 803. Passing through the point (2,40)

( )1 x

cf tab

=+

80c =(0) 16f =

(2) 40f =

8080( )1 xf t

ab=

+

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0

80(0) 161

fab

= =+

80161 a

=+

80

2

80(2) 401 4

fb

= =+

2 801 440

b+ =

280116

a+ =

4a =

24 1b =12

b =

( )80( )

1 4 0.5 tf t =+

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I. Inverses of Exponential functionsA. If , then . ( )( ) where 0, 1xf x b b b= > ≠ 1( ) logbf x x− =

8

6

4

y=x

y=logbx

y=bx

1 b<8

6

4

y=x

y=bx

0 1b< <

2

-2

-4

-6

-5 5

2

-2

-4

-6

-5 5

y=logbx

B. Changing between logarithmic and exponential formIf x > 0 and 0 < b, and b is not equal to 1, then

log yby x b x= ⇔ =

C. Example 1: Evaluate the following logarithms

31. log 9

212. log

323

73. log 7

22 since 3 9= =

1331 since 7 7= =

5 15 since 232

−= − =

73. og 7

184. log 1

125. log 12

00 since 18 1= =

since 7 73

= =

11 since 12 12= =

56. log 0 since 12 0x=∅ ≠

D. Basic Properties of LogarithmsIf x > 0 and 0 < b, and b is not equal to 1, then

0

1

log 1 0 since 1

log 1 since

log since

b

b

y y yb

b

b b b

b y b b

= =

= =

= =log since log logb

b

xb bb x x x= =

E. Example 2: evaluate the following expressions

F. Example 3: Find two integers m and n such that

41. log 643

32. log 97log 83. 7

34log 4 3= =

23 2 3

3 32log 3 log 33

= = =

8=

l 1343log 134m n< <81 134 243< <

4 53 134 3< <4 5

3 3log 3 134 log 3< <

34 log 134 5< <4, 5m n= =

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II. Common Logarithms – Base10A. Logarithms with base 10 are called common

logarithms.B. When writing a log, if we drop the base, it is

assumed to be 10

C Definition:

10log logx x=

C. Definition:

log 10yy x x= ⇔ =

D. Example 4: Evaluate the following:

41. log 10

2. log 0.01

14 1log10

4= =

2log10 2−= = −

log113. 10

4. log10,000 4log10 4= =

11=

E. Example 5: Evaluate the following with a calculator.

1. log84.3

2. log 0.15

1.9261.926 since 10 84.3≈ ≈

0.8230.823 since 10 0.15−≈ − ≈

( )3. log 43.7− since 10 43.7y=∅ ≠ −

F. Example 6: Solve each of the following.

3. log 1x = −4

14. log2

x =

1. log 4x = 52. log 3x =410 10,000x = = 35 125x = =

1 11010

x −= =1

5. log 8 3b = 36. log 272b = −

10 124 2x = =

3

13

8

8 2

b

b

=

= =

32

23

27

1279

b

b

−

−

=

= =

III. Natural Logarithms – base e.A. Logarithms with the natural base e are natural

logarithms.B. These are written with the abbreviation ln for

natural logs

C. Definition

log lne x x=

ln yy x e x= ⇔ =

D. Example 7: Evaluate each of the following without your calculator 81. ln e ( )82. ln e

53. ln e 4. ln 0

8log 8e e= = ( )8 8log 1 1e e= = =

15 1log

5e e= = since 0xe=∅ ≠

ln135. e 13 since ln13 ln13= =

E. Example 7: Evaluate each of the following with your calculator

1. ln8

2. ln 2

2.0792.079 since 8e≈ ≈

0.6930.693 since 2e≈ ≈

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IV. Measuring Sound Using DecibelsA. Definition of Decibels

1. The level of sound intensity in decibels (dB) is

2. (beta) is the number of decibels3. is the sound intensity in W/m2 (Watts per square

meter)h h h ld f h h h d bl

βI

0

10log II

β⎛ ⎞

= ⎜ ⎟⎝ ⎠

4. The threshold of human hearing (the quietest audible sound intensity is )

1220 10 WI m

−=

B. Example 9: Answer each of the given questions. 1. Determine the number of decibels in a human

conversation if the intensity of the sound of a human voice is 10-6 watts per square meter.

6

12

1010log10

β−

−

⎛ ⎞= ⎜ ⎟

⎝ ⎠( )6 ( 12)10l 10β ( )6 ( 12)10log 10β − − −=

610log10β =10 6β = ⋅

60β =

2. An ordinance bans noise of more than 80 decibels after midnight. What is the intensity of an 80 decibel sound?

1280 10log10

I−

⎛ ⎞= ⎜ ⎟⎝ ⎠

128 log10

I−

⎛ ⎞= ⎜ ⎟⎝ ⎠

I81210

10I−=

8 1210 10 I−⋅ =410 I− =

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I. Properties of LogarithmsA. Since logarithms are exponents, they follow rules just 

like exponents.B. Properties of Logarithms

Properties of Logarithms

Let b, R, and S be positive real numbers where b is not equal to 1, and c is any real numbernumber

Product Rule

Quotient Rule

Power Rule

( )log log logb b bRS R S= +

log log logb b bR R SS= −

log logcb bR c R=

C. Example1: write the following expressions as a single logarithm.

5 5 511. 3log 4log log2

w x y+ −1

3 4 25 5 5log log logw x y+ −

3 45 5log logw x y−

3 4

5log w xy

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

5 5 5g g g y

322. log loga a+

1 32 2log loga a+

1 32 2log a a

⎛ ⎞⋅⎜ ⎟

⎝ ⎠

2

2 23. log logx yy z−

2

2log x yy z

⎛ ⎞÷⎜ ⎟

⎝ ⎠

2 2log x z⎛ ⎞⋅⎜ ⎟⎝ ⎠

2log a2log a

2 2logy y⎜ ⎟

⎝ ⎠

2 3log xzy

⎛ ⎞⎜ ⎟⎝ ⎠

D. Example 3: Express the given logarithm in terms of log(x), log(y), and log(z).

4

21. log x yz

⎛ ⎞⎜ ⎟⎝ ⎠

4 2log log logx y z= + −4log log 2logx y z= + −

2 l x12x⎛ ⎞ 1 x⎛ ⎞

2 22. logy z− 2 2log x

y z−

⎛ ⎞= ⎜ ⎟

⎝ ⎠2 2

1 log2

xy z−

⎛ ⎞= ⎜ ⎟

⎝ ⎠2 21 log log log

2x y z−⎡ ⎤= − −⎣ ⎦

[ ]1 log 2log 2log2

x y z= − +

1 log log log2

x y z= − +

3 53. log x yz ( )1

1 23 5 2log x yz

⎡ ⎤= ⎢ ⎥

⎣ ⎦( )

3 152 4log x yz

⎡ ⎤= ⎢ ⎥

⎣ ⎦

3 1 52 4 4log x y z

⎡ ⎤= ⎢ ⎥

⎣ ⎦

3 1 5log log log2 4 4

x y z= + +

II. Change of BaseA. Example 4: Solve the equation y=log612.

6log 12y =6 12y =

ln 6 ln12y =ln 6 ln12y =

ln12ln 6

y = 1.387y ≈

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B. This demonstrates the change‐of‐base formula:

1. For positive real numbers a, b, and x where a and bare not 1: loglog

loga

ba

xxb

=

2. As a result of the buttons on the calculators, we generally use one of the following expressions:

log lnloglog lnb

x xxb b

= =

3. Example 5: Evaluate each of the following logarithms

100a. log 101

2

log10 log10 1log100 log10 2

= = =

18b. log 5 ln5 0.557ln18

= ≈

2

1ln ln1 ln 3 0 ln 3 ln 3 13ln 9 ln 9 ln 3 2ln 3 2

− − −= = = = = −9

1b. log3

⎛ ⎞⎜ ⎟⎝ ⎠

III. Graphs of Logarithmic Functions with Base b.A. Any logarithmic function                    can be written as 

a transformation on the natural log function.( ) logbg x x=

ln 1( ) log lnln lnb

xg x x xb b

= = =

2

-2

-4

5

f x( ) = log x( )

log 3( )

B. Example 4: describe how to transform the graph of f(x)=ln(x) into the graph of the given function, then sketch the given function. 

71. ( ) logg x x= 1 lnln 7

x=

1vertical shrink by 0.514ln 7

≈

2

-2

5 10

g(x)

f(x)

13

2. ( ) logh x x=( ) 1

1 1 1ln ln ln1 ln3 ln3ln 3

x x x−= = =−

1reflect over the x-axis and vertical shrink by 0.91ln 3 ≈

2

-2

5 10

y=h(x)

y=f(x)

IV. Re‐expressing DataA. Sometimes we can find information about bi‐variate

data by applying a function to one or both variables in the data set.

B. The data below represents the average length and weight at different ages for Atlantic Ocean rockfish

    8 6

1. First, type this data into your calculator, and find the log of both sets of data (you must remove the (0,0) point (why?).

x: age (years)

0 4 8 12 16 20

weight (grams) 

0 48 192 432 768 1200

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2. Now, find a linear relation between these variables.

3. We will finally return to the original variables as follows:

ln 2ln 1.099y x= +

2 1.099ln ln lny x e≈ +

( )2 1.099ln lny x e≈ ⋅( )2 1 099l

4. Compare this answer to your answer using the power regression (of the original data) on your calculator.

( )2 1.099lnln x eye e ⋅≈

2 1.099y x e≈23.001y x≈

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3.5: Equation Solving and Modeling

I. Solving Exponential Equations

A. One-to-One Properties:

1. For any exponential function f(x)=bx,

u vb b u v= ⇔ =2. For any logarithmic function f(x)=logbx,

b b u v= ⇔ =

log logb bu v u v= ⇔ =

B. Example 1: solve the following exponential equation.

2

1108 43

x⎛ ⎞ =⎜ ⎟⎝ ⎠

2

1 1x⎛ ⎞

2 3x=

21 13 27

x⎛ ⎞ =⎜ ⎟⎝ ⎠

2 31 13 3

x⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

23

x=

C. The function sinh(x) (pronounced “sinch x”) is called the hyperbolic sine of x, is defined by the equation.

D. Example 2: Determine all values for which sinh(x)=1/4.

sinh( )2

x xe ex−−

=

1x xe e−−( ) 12 4

e e=

( ) ( ) 14 42 4

x xx xe ee e

−⎛ ⎞− ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

22 2x xe e− =22 2 0x xe e− − =

( ) ( )( )( )

21 1 4 2 22 2

xe± − − −

=

1 174

xe ±=

use the quadratic formula to solve for ex

now we take the natural log of both sides and remove the negative solution (why?)

( ) 1 17ln ln4

xe⎛ ⎞+

= ⎜ ⎟⎜ ⎟⎝ ⎠

0.2475x ≈

E. Example 3: Solve the following logarithmic equation.

23log 12x =method 1

2log 4x =2 10 000x =

method 22log 4x =

2 410x =10,000x =100x = ±

both answers work, because -100becomes positive when substituted

10x210x = ±

100x = ±

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method 3 (wrong)2log 4x =

2log 4x =log 2x =

210x =100100x =

How and why did this fail?it missed the negative solutionthe power rule only works for positive values of Ryou should check your answers with your calculator

F. Example 4: Solve the following logarithmic equation

log( 3) log(2 1) 2log 2x x+ + − =

[ ]log ( 3)(2 1) log 4x x+ − =( 3)(2 1) 4x x+ − =

22 5 7 022 5 7 0x x+ − =

( 1)(2 7) 0x x− + =71,2

x = −

note: since -7/2 makes (at least one of) x-3 and 2x-1 negative,

the logs are undefined. Therefore the only solution is: 1x =

II. Orders of Magnitude

A. The common logarithm of a positive quantity is its order of magnitude.

B. This is useful for comparing quantities with huge variability, such as distances to planets or sizes of elementary particleselementary particles.

C. Example 5: compare the order of magnitudes for the mass of an electron to that of a proton. The mass of an electron is approximately 9.11E-31, while the mass of a proton is approximately 1.67E-27 kg.

1. We can estimate by rounding the masses as follows:

2 We can also use the calculator to find slightly better

31 31 30

27 27 27

electron: 9.11 10 10 10 10proton: 1.67 10 1 10 10

− − −

− − −

⋅ ≈ ⋅ =

⋅ ≈ ⋅ =

3

thus, the proton is 3 orders of magnitude more massive than an electronor in other words, it is 10 1000 times larger=

2. We can also use the calculator to find slightly better approximations.

( )( )

31

27

log 9.11 10 30.04

log 1.67 10 26.78

−

−

× ≈ −

× ≈ −thus, the proton is 26.78 ( 30.04) 3.26 orders of magnitude more massive than the electron

≈ − − − =

D. More examples (already rounded):1. The radius of the sun (10^9) is 2 orders of

magnitude longer than the radius of the earth (10^7).

2. The universe (10^55) is 11 orders of magnitude larger than the milky way galaxy (10^44)

3. The population of the world (10^10) is 3 orders of p p ( )magnitude larger than the population of New York City (10^7).

4. A kilometer (10^3) is 12 orders of magnitude larger than a nanometer (10^-9)

IV. Logarithmic ScalesA. Richter scale magnitude

1. a is the amplitude in micrometers of the vertical ground motion at the receiving station

2 T is the period of the associated seismic wave in

log aR BT⎛ ⎞= +⎜ ⎟⎝ ⎠

2. T is the period of the associated seismic wave in seconds

3. B accounts for the weakening of the seismic wave with increasing distance from the epicenter of the earthquake.

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B. Example 6: How many times more severe was the Chinese earthquake ( RC=7.9) than the Pakistani earthquake (RP=7.9), both in 2008?

log 7.9

log 6.4

CC

PP

aR bTaR bT

⎧ = + =⎪⎪⎨⎪ = + =⎪⎩

log log 7 9 6 4C Pa aB B⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟log log 7.9 6.4C PB BT T

+ − + = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

log log 1.5C Pa aT T

− =

log 1.5C Pa aT T

⎛ ⎞÷ =⎜ ⎟⎝ ⎠

log 1.5C

p

aa

⎛ ⎞=⎜ ⎟⎜ ⎟

⎝ ⎠

1.510 31.62 timesC

p

aa

= ≈

C. pH scale

1. The hydrogen concentration is written as [H+].

2. These concentrations generally involve negativepowers of ten hence the (-) in the formula

logpH H +⎡ ⎤= − ⎣ ⎦

powers of ten, hence the ( ) in the formula.3. More acidic solutions have higher hydrogen-ion

concentrations and lower pH values.

D. Example 7: Calculate the pH of each of the following:1. Normal blood: H+ =3.4E-8

2. Ammonia: H+ =1.6E-11( )8log 3.4 10 7.469 /moles liter−− × ≈

( )11log 1.6 10 10.796 /moles liter−− × ≈

E. Example 8: Given the following pH values, find the corresponding hydrogen ion concentrationsp g y g

1. Eggs: pH = 7.8

2. Tomatoes: pH = 4.2

7.8 log H +⎡ ⎤= − ⎣ ⎦7.810H + −⎡ ⎤ =⎣ ⎦

4.2 log H +⎡ ⎤= − ⎣ ⎦ 4.210H + −⎡ ⎤ =⎣ ⎦

F. Example 9: 1. How many times greater is the hydrogen-ion

concentration in tomatoes than eggs?

2. By how many orders of magnitude do the concentrations differ?

III. Newton’s Law of Cooling

4.23.6

7.8

of tomatoes 10 1010 of eggs

H

H

+ −

−+

⎡ ⎤⎣ ⎦ = =⎡ ⎤⎣ ⎦

3.6log10 3.6=

. gA. Newton’s Law: an object that has been heated will cool to the

temperature of the medium in which it is placed.

1. k is a constant that depends on the medium2. Tm = the temperature of the surrounding medium.3. T0 = the initial temperature of the object.

( )0( ) ktm mT t T T T −= + −

B. Example 10: You cook a pizza at 350 degrees Fahrenheit , and then let it cool in a room that is 68 degrees Fahrenheit. After 10 minutes, the pizza has cooled to 240 degrees.

1. Find a specific equation.

0 0350, 68, 282, (10) 240m mT T T T T= = − = =( )0( ) kt

m mT t T T T e−= + −

( )( ) 68 282 ktT t e−= +( ) ( )10k

( )100.6099 ke−≈( ) ( )ln 0.6099 10k≈ −

( ) ( )10(10) 240 68 282 kT e−= = +

( )0.4944 10k− ≈ −0.04944 k≈

( ) 0.0494( ) 68 282 tT t e−= +

2. What is the temperature of the pizza after 15 minutes?

3. After how many minutes will the pizza be 150 degrees?

(15) 202.41T C≈

( ) ( )0.0494 15(15) 68 282T e−= +

( ) 0.0494( ) 150 68 282 tT t e−= = +0.049440 2908 te−≈0.2908 e≈

( )ln 0.2908 0.0494t= −

25.002 minutest ≈

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C. Modeling with Newton’s Law of CoolingExample 11: The data below shows the cooling temperatures of a cup of coffee after it is poured from the brewing pot into a serving cup. The temperature of the air is 70 degrees Celcius

Time (mins) Temp. (deg C) T-Tm5 168.7 98.7

8 158 1 88 18 158.1 88.1

11 149.2 79.2

15 141.7 71.7

18 134.6 64.6

22 125.4 55.4

25 123.5 53.5

1. Fit an exponential regression model to the T-Tm.

2. Estimate the temperature of the coffee when it was first brewed.

0.031( ) 113.56 tf t e−=

0this is the ' ' part of Newton's law, so:mT T−0.031( ) 70 113.56 tT t e−= +

( )0.031 0(0) 70 113.56T e−= +

3. Estimate how long it would take for the coffee to cool to 100 degrees.

(0) 183.56T C≈

0.031( ) 100 70 113.56 tT t e−= = +0.0310.264 te−≈

ln(0.264) 0.031t≈ −

42.94mint ≈

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1

I. Interest Compounded AnnuallyA. Interest computed annually:

1. Suppose a principal of P dollars is invested in an account earning r interest and calculated annually.

Time in Years Amount in the Account

0

10A P principal= =

1 (1 )A P r= +

2

3

.

.

.

n

1 (1 )A P r+

22 1 (1 ) (1 )A A r P r= ⋅ + = +

33 2 (1 ) (1 )A A r P r= ⋅ + = +

(1 )nnA A P r= = +

.

.

.

2. Formula for annually compounded interest:

3. Example 1: Suppose Morgan invests $750 at 3.5% interest compounded annually. Find the value of her investment 8 years later.

(1 )nA P r= +

her investment 8 years later.

( )8750 1.035 $987.61A = ≈

II. Interest Compounded k times per year.A. Sometimes interest on an investment will be

compounded k times per year (instead of once).

B. Example 2: Suppose Abigail invests $1200 at 6% annual interest compounded monthly. Find the

1ktrA P

k⎛ ⎞= +⎜ ⎟⎝ ⎠

value of her investment 11 years later.

4110.061200 1 $2310.404

A⋅

⎛ ⎞= + ≈⎜ ⎟⎝ ⎠

C. Example 3: Grace has invested $625 at 5.15% interest compounded monthly. How long will it take for her investment to grow to $2500 (graphically)?

120.05152500 625 1t⋅

⎛ ⎞= +⎜ ⎟2500 625 112

+⎜ ⎟⎝ ⎠

27t years≈

D. Example 4: Summer invests $725 in an account that is compounded quarterly. What interest rate is required to double her money in 15 years?

415

1550 725 14r ⋅

⎛ ⎞= +⎜ ⎟⎝ ⎠

60

2 14r⎛ ⎞= +⎜ ⎟

⎝ ⎠⎝ ⎠60

ln 2 ln 14r⎛ ⎞= +⎜ ⎟

⎝ ⎠

ln 2 60ln 14r⎛ ⎞= +⎜ ⎟

⎝ ⎠

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2

ln 2 ln 160 4

r⎛ ⎞= +⎜ ⎟⎝ ⎠

ln 260 1

4re = +

⎛ ⎞ln 2604 1e r

⎛ ⎞− =⎜ ⎟

⎝ ⎠

0.0465r ≈

III. Interest Compounded ContinuouslyA. When interest is ‘always’ being compounded, we

say it is compounded continuously.

lim 1kt

k

rt

rA Pk

A Pe

→∞

⎛ ⎞= +⎜ ⎟⎝ ⎠

=

B. Example 5: Julia invests 2400 in an account that is compounded continuously with a rate of 2.5% for 5 years. How much does she have at the en of the 5 years?

0.025 52400 $2719.56A e ⋅= ≈

IV. Annual Percentage Yield

A. The Annual Percentage Yield (APY) is a basis for comparing investments.

B. Definition: the percentage rate that, compounded annually, would yield the same return as the given interest rate with the given compounding period.

C. Example 6: Sue invests $1850 at a bank at 6.25% annual interest compounded monthly. What is the equivalent APY? (this is per year, so we use 1 for t)

( )120.06251850 1 1850 1

12x ⎛ ⎞+ = +⎜ ⎟

⎝ ⎠

( )120.06251 1

12x ⎛ ⎞+ = +⎜ ⎟

⎝ ⎠( )

12⎜ ⎟⎝ ⎠

120.06251 112

x ⎛ ⎞= + −⎜ ⎟⎝ ⎠

0.0643x ≈

D. Example 7: Which is more attractive, one that pays 6.2% compounded continuously or 6.25% compounded monthly?

1

2

is the APYfor the 6.2% rate is the APYfor the 6.25% rate

rr

0.06211 r e+ =

0 062 1

12

20.06251 1

12r ⎛ ⎞+ = +⎜ ⎟

⎝ ⎠0.0621 1r e= −

1 0.0640r =

⎝ ⎠12

20.06251 1

12r ⎛ ⎞= + −⎜ ⎟

⎝ ⎠

2 0.0643r =

, the 6.25% rate is better∴

V. Annuities—Future Value

A. An annuity is a sequence of equal periodic payments.

B. It is an ordinary annuity if deposits are made at the end of each period at the same time the interest is posted in the account.

C. The total value of the investment returned form an annuity including payments and interest is called the future value.

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D. Future Value of an Annuity.

1. n is the number of equal periodic payments

( )1 1niFV R

i+ −

=

2. R is the amount of the payments

3. i is the interest rate per compounding period (payment interval)

E. Example 8: At the end of each month, Courtney pays $250 into a mutual fund. If her investments earn 6.75% annual interest compounded monthly, what will be the value of her annuity in 30 years?

• R=250( )1 1ni+

• i=0.0675/12

• N=12(30)

( )1 1iFV R

i+ −

=

12 300.06751 112250 0.0675

12

FV

⋅⎛ ⎞+ −⎜ ⎟⎝ ⎠=

( )3601 0.005625 1250FV

+ −=

( )2500.005625

FV =

$290,366.47FV =

VI. Loans and Mortgages—Present ValueA. The net amount of money put into an annuity is

its present value.B. The net amount returned from the annuity is its

future value.C. Present Value of an Annuity

( )1 1 niPV R

−− +=

1. n is the number of equal periodic payments2. R is the amount of the payments3. i is the interest rate per compounding period

(payment interval)

PV Ri

D. Example 9: Chris purchases a new car for $25,000. What are the monthly payments for a 5 year loan with a $3000 down payment if the APR (Annual Percentage Rate) is 3.2%?

The amount borrowed is actually $22,000, therefore:

PV = $22,000i = 0.032/12You are solving for the payment R.

( )1 1 niPV R

i

−− +=

( )5 120.0321 11222000 0.032

12

R

−⎛ ⎞− +⎜ ⎟⎝ ⎠=

( ) 601 1 0.0026722000 R

−− +=22000

0.00267R

22000 55.378R=

$397.27R ≈