R—05/28/09—HW #73: Pg 477:47,49,50; Pg 490:17,49-61odd; Pg 496:31-55 eoo; Pg 505:25-59 odd50) V=22000(.875)^t; 14,738

F—05/29/09—HW #74: Pg 469: 19-24; Pg 477: 19-24; Pg 496: 33-57 eoo; Pg 505: 26-58 even20) E 22) A 24) D

20) E 22) B 24) A

26) -1\5 28) no sol 30) ln(1\6) 32) (ln 3)\534) ln 5\3 36) -.661 38) (ln5.5)\4 40) ln13.542) log255 44) 1\2 46) e^(15\8) 48) e^(5\2)50) 15 52) .729 54) 256 56) no sol58) no sol

Chapter 8 Review

xy 2x

y

2

1

An exponential function involves y = bx where the base b is positive and not equal to 1.

An asymptote is a line the graph approaches.

Usually represented by a dotted line

An exponential growth function is when b > 1

An exponential decay function is when 0 < b < 1

Tell me if the following are growth or decay functions.

xy 2x

y

2

12

xy 24x

y

3

5

6

1

Decay Growth Growth

x

y

2

14 Decay

One to Review

1) Find key points

2) Add ‘h’ to the x-value

3) Multiply ‘a’ to the y-value

4) Add ‘k’ to the y-value

242 1 xy )4,1()1,0(4

1,1

Other info to find:

y – intercept:

Asymptote:

Domain: All real numbers

Range:

End behavior: x ∞; f(x) _____

x -∞; f(x) _____

h = 1)4,2()1,1(

4

1,0

a = 2 )8,2()2,1(2

1,0

k = -2)6,2()0,1(

2

3,0

2

3242 1

2y

),2(;2 y2

A town of 5000 grows at a rate of 10% per year. How many people are in town after 10

years?

y = a(1 + r)t

y final value

a starting value

r % increase, in decimal form

(1 + r) Growth Factor

t Time

Find the info

Write equation

Answer question

5000

.1

1.1

10

5000

.1

1.1

10

y = (1 + )Calculator Plug – In

5000 * (1.1)^10

12968 people

(answers must make sense)

A car that costs 25,000 depreciates at a rate of 5% per year.

A) What is the value of the car after 7 years?

B) When is the car worth $20,000?

Calculator – Windows and Intersect

7)^95(.25000

)05.1(25000 7y

43.17458$

y = a(1 – r)t

a

r

Decay factor

t

25000

.05

.95

7

20000

)95(.25000

y

y x

You can treat it like a number.

23 ee 23e

5e

xe

e5

3

xe 53

43 )( xe

43 xe

xe12

13 )2( xe)1(31)2( xe

22

33 xx ee

3 6xe3 32 )( xe

xe2

Let b and y be positive numbers, and b = 1. The LOGARITHM of y with base b is logby and is defined

as

logby = x if and only if bx = y

You read this log base b of y equals x.

log 125 35

log

22

1

4 125 35

22

1

4

Rewrite in Exponential Form

01log5 14log4 29log3

1 2

532log4

4log2isWhat

?42 equalspowerwhatto

2

8log2

1isWhat

?82

1equalspowerwhatto

3

100log10isWhat

?10010 equalspowerwhatto

2

10

10

loglog

logarithmcommon

10

log,10

log

means

acalledisBase

x

baseisbuttonthe

calculatoryourOn

e

e

means

acallediseBase

xe

baseisbuttonthe

calculatoryourOn

logln

logarithmnatural

log,

ln

54.4ln

ln

20log

e

3010.1

1

5129.1

Log Properties: g(x) = logbx is the inverse of f(x) = bx

That means g(f(x)) = logbbx = x, and f(g(x)) = xb xb log

Try to make things match up with the base, and it’ll work out ok.

x10log10x3log3

xe 2ln x27log3

27

1log3isWhat

e

Calc

log

Finding inverses

A) Simplify first

1) Switch x and y

2) Change forms

3) Simplify

)3ln( xyy x

)3ln( e

3

yex 3

yx 3log yx )4log(

Finding inverses

A) Simplify first

1) Switch x and y

2) Change forms

3) Simplify

6021.4log 0792.112log

48log

124log

12log4log

6813.1

0792.16021.

nun

vuv

u

vuuv

rules

bu

b

bbb

bbb

loglog

logloglog

logloglog

!log

3log

4

12log

4log12log

4771.

6021.0792.1

3

1log

12

4log

12log4log

4771.

0792.16021.

16log

24log

4log2

2042.1

)6021(.2

144log

212log

12log2

1584.2

)0792.1(2

!log ruleswithSolve

24log x

2log4log x

xlog24log

!expandorCondense

2log10log

2

10log

onenttoroot

changeClue

exp

,5log

3

10log

15log3log2log x )log210(log6log2

1y

2

1

5log

5log2

13log10log

5log

15log3loglog 2 x15log9loglog x

15log9log x

15

9log

x

5

3log

x

)log10(log6log 22

1

y

)10(log6log 22

1

y

2

2

1

10

6log

y

3log1

• General Solving Rules• Methods of solving

– Make bases of exponents the same• Notice when both sides have x as an exponent, and it looks like you

can make bases the same.

– Log both sides• Generally if you have a variable exponent on only one side.• Make the base of your log the same as the base of the exponent

– Make terms inside logs equal• Both sides of the equation have logs with same bases• May involve condensing log expressions

– Exponentiating both sides• One side has a log, one side doesn’t• May involve condensing log expressions

– NEED TO DOUBLE CHECK!!!!

x225

!exp

,

samethebemustonentsso

sametheisBase

x5

41 22 x

41x

3x

124 xx

12 xx

1x

12 2)2( xx

61 24 x

622 x

42 x

6)1(2 2)2( x

2x

83255

1 xx

235 162 xx

7

16

7

8

64 5 x

636 x

922 x

72 x

1427 3 x

8074.2

2925.6

2263.1

5850.1

7230.1

)4(log)2(log 33 xx xx 42 x32

x3

2

)10ln()2ln( xx

102 xx

10x

)2(log)32(log 66 xx

232 xx

1x

)5(log)63(log 33 xx xx 563 x26 x 3

CAN’T HAVE NEGATIVE INSIDE OF LOGS! EXTRANEOUS SOLUTION.

MUST ALWAYS DOUBLE CHECK!!!!!

Solution

NO

2log3 x

2log 33 3 x

23x

9x

9)2(log3 4 x

Convert 3)2(log 44 4 x

342 x66x

3)2(log4 xMUST ALWAYS DOUBLE CHECK!!!!!

Exponentiating both sides

2)1(log)2(log 22 xx

2))1)(2((log2 xx

062 xx

2,3

0)2)(3(

xx

xx

2)ln()1ln( xx

2)ln( 2 xx

22 exx

)1(2

))(1(4)1()1( 22 ex

2)2(log 22 xx

2)2(log 222

2 xx

422 xx

2)ln( 2

ee xx

022 exxwork

tdoesnanswerOne

eApproximat

'

formula

quadraticntHi ,

MUST ALWAYS DOUBLE CHECK!!!!!

• R—05/28/09—HW #73: Pg 477:47,49,50; Pg 490:17,49-61odd; Pg 496:31-55 eoo; Pg 505:25-59 odd

• F—05/29/09—HW #74: Pg 469: 19-24; Pg 477: 19-24; Pg 496: 33-57 eoo; Pg 505: 26-58 even

•