Exponential and Log

8/23/2019 Exponential and Log

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ExponentialandLogarithmicFunctions


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Exponential Function

An exponential function

y = f(x) = bx where b>0 and bcannot be equal to 1.

The domain of the exponential function with

base b is the set of real numbers. Its rangeis the set of positive real numbers.


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Properties of Exponents

Ifb is any real number and n is any positive

integer, then

1. If n = 1, bn

= b2. if n > 1, b n = b b n -1

3. if b 0, b n =

4. if b 0, b0 = 1

nb

1


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Exponential Equations

An equation in which the unknown is an exponent is called an

exponential equation.For example, 2x = 16, 3x + 2= 81

To solve an exponential equation, write both sides as

exponential expressions with the same base. If the

expressions are equal and the bases are equal, then

the exponents must be equal.

This method can be used because the exponential

function is one to one.


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Example # 1

Solve for b: 9 = b 2 / 3

Solution

9 = b 2 / 3

9(3/ 2) = b2/ 3 (3/ 2)

3 3 = b1

27 = b


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Example # 2

Solve for x: 125 = 5x

Solution:125 = 5x

53 = 5x

3 = x


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GraphingExponential

Functions


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Basic Properties of the Graph ofExponential Function

1. All graphs pass through the point (0, 1).

b0 = 1 for any permissible base b.

2. All graphs are continuous, with no holes or

jumps.

3. The x axis is the horizontal asymptote.

4. If b > 0, then bx increases as x increases.

5. If 0


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With your Graphing Calculatorgraph each of the following

y = 2x

y = 3x

y = 5xy = 1x

Determine what is happening when thebase is changing in each of these graphs.


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y = 2x

x y = 2x y = 3x

-2 1/4 1/9

-1 1/3

0 1 1

1 2 3

2 4 9

3 8 27

y = 3x


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y = 2x

x y = 5x y = 1x

-2 1/25 1

-1 1/5 1

0 1 1

1 5 1

2 25 1

3 125 1

y = 3xy = 5x

y = 1x


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y = 2x

y = 3xy = 5x

y = 1x

Determinewhere each ofthe followingwould lie?

y=10x

y=4x

y = (3/2)x

y = 10xy = 4x

y = (3/2)x


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We also discovered thatchanges in a would

change the y-intercept on itscorresponding graph.

Now lets turn our attention

to a useful property ofexponential functions.

xy a b


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f(x) = 2x


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f(x) = 2x-3

f( ) 2 +2


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f(x) = 2x+2 -3


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f(x) = -(2)x-4 2


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With your Graphing Calculatorgraph each of the following

y = 1x

y = (1/2)x

y = (1/3)x

Determine what is happening when thebase is changing in each of these graphs.


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x y = logx

1/4 2

1

1 0

2 -1

4 -2

8 -3

y = logx

x = ()y

y = 5xy = (1/3)x


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y = 2x

x y = ()x

y = (1/3)x

-2 4 9

-1 2 3

0 1 1

1 1/3

2 1/9

3 1/8 1/27

y = 3xy = 5x

y = 1x

y = (1/3)x

y = ()x


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f(x) = ()x-3 - 2 = (2)-x+3 - 2


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f(x) = 2-x = (1/2)x


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A new Number

718.20

!1

ne


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A new Number

!5

1!4

1!3

1!2

1!1

1!0

1e

We could use a spreadsheet to determine an approximation.

120

1241

61

21

11

11e

0

!1n

e

3x


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y = 2x

x y = 2x

y = 3x

-2 1/9

-1 1/3

0 1 1

1 2 3

2 4 9

3 8 27

y = 3x

y = exGraph y = ex


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y = exy = ex+2Graph:

y = ex+2

x + 2 = 0

x = -2


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If a quantity increases by the same proportion r in each unit

of time, then the quantity displays exponential growth and

can be modeled by the equation

y C r

t ( )

1

Where

C = initial amount

r = growth rate (percent written as a decimal)

t = time where t 0(1+r) = growth factor where 1 + r > 1


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You deposit $1500 in an account that pays 2.3% interest compounded yearly,

1) What was the initial principal (P) invested?

2) What is the growth rate (r)? The growth factor?

3) Using the equation A = P(1+r)t, how much money would you have after

2 years if you didnt deposit any more money?

3) A P r

A

A

t

( )

( . )

$ .

1

1500 1 0 023

1569 79

2

1) The initial principal (P) is $1500.

2) The growth rate (r) is 0.023. The growth factor is 1.023.


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You deposit $1500 in an account that pays 2.3% interest compounded yearly,

1) What was the initial principal (P) invested?

2) What is the growth rate (r)? The growth factor?

3) Using the equation A = P(1+r)t, how much money would you have after

2 years if you didnt deposit any more money?

3) A P rA

A

t

( )( . )

$ .

1

1500 1 0 023

1569 79

2

1) The initial principal (P) is $1500.

2) The growth rate (r) is 0.023. The growth factor is 1.023.


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Compound Interest

nt

nr

PA 1

You deposit $5000 into an account that pays 4.5 %interest. What is the balance of the account after10 years if the interest is compounded quarterly?

A = Final amount = unknown

P = Principal = $5000r = rate of interest = .045

n = number of times compounded per year = 4

t = number of years compounded = 10


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Compound Interest

nt

nr

PA 1


A = unknown

P = $5000r= .045

n = 4

1044045.015000

A

t = 10

88.7821$A


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Compound Interest

nt

nr

PA 1


A = unknown

P = $5000r= .045

n = 4

105252045.015000

A

t = 10

04.7840$A

weekly?

52


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Compound Interest-continuously

rt

PeA

You deposit $5000 into an account that pays 4.5 %interest. What is the balance of the account after 10years if the interest is compounded continuously?

A = Final amount = unknown

P = Principal = $5000r = rate of interest = .045

t = number of years compounded = 10


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Compound Interest-continuously

rt

PeA

You deposit $5000 into an account that pays 4.5 %interest. What is the balance of the account after 10years if the interest is compounded continuously?

A = unknown

P = $5000r= .045

t = 10

10045.05000 eA

56.7841$A


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Bacteria Growth

kt

ney

You have 150 bacteria in a dish. It the constant ofgrowth is 1.567 when t is measured in hours. Howmany bacteria will you have in 7 hours?

y = Final amount = unknown

n = initial amount = 150k = constant of growth = 1.567

t = time = 7


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Bacteria Growth

kt

ney

You have 150 bacteria in a dish. It the constant ofgrowth is 1.567 when t is measured in hours. Howmany bacteria will you have in 7 hours?

y = unknown

n = 150k = 1.567

t = 7

7567.3150 ey

678.977,706,8ybacteria678.977,706,8


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If a quantity decreases by the same proportion r in each unit

of time, then the quantity displays exponential decay and

can be modeled by the equation

y C r t ( )1

Where

C = initial amount

r = growth rate (percent written as a decimal)

t = time where t 0(1 - r) = decay factor where 1 - r < 1


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You buy a new car for $22,500. The car depreciates at the rate

of 7% per year,

1) What was the initial amount invested?

2) What is the decay rate? The decay factor?

3) What will the car be worth after the first year? The second

year?

1) The initial investment was $22,500.

2) The decay rate is 0.07. The decay factor is 0.93.

3 1

22 500 1 0 07

20 925

1

) ( ), ( . )

$ ,

y C ry

y

t

y C ry

y

t

( ), ( . )

$ .

1

22 500 1 0 07

19460 25

2


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y = 2x

x y

-2 1/4

-1

0 1

1 2

2 4

3 8

x y

1/4 -2

-1

1 0

2 1

4 2

8 3

x = 2y


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How do we

solve this

exponential

equationfor the variable y?

y = 2x x = 2y


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y = 2x

x y

-2 1/4

-1

0 1

1 2

2 4

3 8

x y

1/4 -2

-1

1 0

2 1

4 2

8 3

x = 2yy=log2x


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LOGARITHMS

exponential logarithmic

b > 0A > 0

Ab

m

mAb

)(log


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2)9(log3 932

exponential logarithmic

3)125(log 5 12553

3log81

2 8132

5)32(log21 32

5

21

Ab

m mAb

)(log

yx 2 yx )(log 2


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Properties of Logarithms

Product Property

Quotient Property

Power Property Property of Equality


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Change of Base Formula

axxa

10

10

logloglog

a

xx

b

ba

log

loglog


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Product Property

nmnm aaa

)(log)(log)(log nmnm bbb

multiplication addition

multiplication addition


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Product Property

)4(log)16(log)416(log 222

)2(log)2(log)22(log 2

2

4

2

24

2

24)2(log 62

66


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Quotient Property

nm

n

m

a

a

a

)(log)(log)(log nm bbnm

b

division subtraction

division subtraction


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Quotient Property

)4(log)32(loglog 22432

2

)2(log)2(log8log 22

5

22

25)2(log 32

33


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Power Property

nmnm aa

logb(mp)

logb(mp) = plogb(m)

p


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Power Property

)2(log72log 27

2

177

77


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Property of Equality

CAthen

)(log)(log CAif bb


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u)25(log 5

Evaluate

255 u

255 u

2u

2)25(log 5


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u)81(log 3

Evaluate

813 u

433 u

4u

4)81(log 3


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u3212log

Evaluate

3212 u

52

12 u

5u

5log321

2

522 u


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u)7(log 7

Evaluate

77 u

177 u

1u

1)7(log 7


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unn )(log5

Evaluate

5nnu

5u

5)(log

5 nn


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u)1(log8

Evaluate

18 u

088 u

0u

0)1(log8


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Evaluate

log3(25) = u

3u = 25

3u = 52

??????


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Solve for x

log2(x+5) = 4

24 = x + 5

16 = x + 5

11 = x


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Solve for x

logx(32) = 5

x5 = 32

x5 = 25

x = 2


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Evaluate

log5(568)

= 3.941

5log568log

10

10


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Evaluate

log3(25)

= 2.930

3log25log

10

10


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xy = log2x

1/4 -2

-1

1 0

2 1

4 2

8 3

y = log2x

y = log3x

y = log5x

x = 2y


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)(log 235 yx

Expand

)(log)(log 253

5 yx

)(log2)(log3 55 yx

product property

power property


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Expand

)(log)(log 4535

5 zyx

)(log)(log)(log 453

5

5

5 zyx

quotient property

product property

)(log4)(log3)(log5 555 zyx power property

4

35

5log zyx


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)(log)(log)(log 552

5

7

5 zyx

)(log)(log 5257

5 zyx

Expand

quotient property

product property

)(log5)(log2)(log7 555 zyx power property

52

7

5log zyx

)(log)(log)(log

5

5

2

5

7

5 zyxdistributive property


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zyx 333 log2log6log5

Condense

power property

product property

2

3

6

3

5

3 logloglog zyx

2

3

65

3 loglog zyx

2

65

3log zyx

quotient property


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410

21010 logloglog

21

zyx

zyx 10101021 log4log2log

Condense

group / factor

product property

41021010 logloglog 21

zyx

421010 loglog 21

zyx

42

21

10log zyxquotient property

Power property

4210log zyx


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4523 loglogloglog wzyx eeee

4253 loglogloglog wyzx eeee

wzyx eeee log4log5log2log3

Condense

re-organize

group

4253

loglogloglog wyzx eeee

4253

loglog wyzx ee

42

53

logwy

zxe

product property

Power property

quotient property


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Solve for x

393 xx

122 x

6x

3log93log 33 xx

Property ofEquality


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3log93log 33 xx

Solve for x

6xcheck

36log9)6(3log 33 36log918log 33

9log9log 33 checks!


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3log93log 33 xx

Solve for x

393 xx

122 x

6x

6


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nn 6log2log7log 444

Solve for n

nn 6147

14n

nn 6log)2(7log 44 Condenseleft side

Property ofEquality


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nn 6log2log7log 444

Solve for n

14ncheck

)14(6log214log7log 444 84log12log7log 444

84log)12(7log

44

84log84log 44

checks!


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nn 6log)2(7log 44

nn 6log2log7log 444

Solve for n

nn 6147

14n

14

S f


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Solve for x

31log1log 22 xx

)1)(1(23 xx

18 2 x2

9 xx 3

3)1)(1(log 2 xxCondenseleft side

Convert toexponential

form

S l f


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Solve for x

31log1log 22 xx

3xcheck 3xcheck

313log13log 22

32log4log 22

312

33

checks!

313log13log 22

34log2log 22

fails

The argumentmust be positive

S l f


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Solve for x

3)1)(1(log 2 xx

)1)(1(23 xx

18 2 x2

9 xx 3 3

31log1log 22 xx

S l f


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Solve for x

33 22 xx

39 2 xx

60 2 xx

)2)(3(0

xx

23log 23 xx

23 xorx

Convert toexponential

form

S l f


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23log 23 xx

Solve for x

3xcheck

checks!

23)3()3(log 23 2339log 3

29log3

22

2xcheck

232)2(log 23 2324log 3

29log3

22

checks!

S l f


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23log 23 xx

Solve for x

33 22 xx

39 2 xx

60 2 xx

)2)(3(0

xx 23 xorx 2,3

S l f


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Solve for x

)19log()5log()73( x

)19log()5log(7)5log(3 x)5log(7)19log()5log(3 x

)5log(3

)5log(7)19log( x

943.2x

19log5log73

x

)5log(3)5log(3

S l f


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24

7log5log x

Solve for x

)7log()2()5log()4( x

)5log(4

)7log(2x

605.0x

)5log(4)5log(4

S l f


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Solve for x

)9log()5()11log()12( xx

)9log(5)11log(1)11log(2 xx )11log()9log(5)11log(2 xx

)11log()9log(5)11log(2 x

)9log(5)11log(2)11log(

x

3870x

xx 512

9log11log

S l f


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Solve for x

)5log()1()3log()2( xx

)5log(1)5log()3log(2)3log( xx)3log(2)5log()5log()3log( xx

)3log(2)5log()5log()3log( x )5log()3log(

)3log(2)5log(

x

12

5log3log

xx

151.1x

S l f


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Solve for x

)3log()89()7log()23( xx

)3log(8)3log(9)7log(2)7log(3 xx)3log(8)7log(2)3log(9)7log(3 xx

)3log(8)7log(2)3log(9)7log(3 x )3log(9)7log(3

)3log(8)7log(2

x

209.1x

8923

3log7log

xx

S l f


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Solve for x

)ln()23()15ln( ex

23)15ln( xx32)15ln(

x

3

2)15ln(

x569.1

23

ln15ln

x

e1

S l f


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Solve for x

)ln()65()ln( 37

ex 65)ln(

37 x

x56)ln(3

7

x

5

6)ln(37

x 031.1

1

65

37

x

e 65

37 lnln xe

Solve for x


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)2log()13()log( 75

x

Solve for x

)2log(3)2log()log(7

5 x

x

)2log(3

)2log()log(75

1 20

13

275

x

1375 2loglog x

)2log(1)2log(3)log(75 x

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Exponential and Log

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