Exponential & Logarithmic Functions

REVIEW11.1 – 11.4

Properties of Exponents

€

am • an = am+ n

€

am

an = am −n

€

(am )n = am• n

€

(a • b)m = am • bm

€

a

b

⎛

⎝ ⎜

⎞

⎠ ⎟m

=am

bm

€

a0 =1

€

a−n =1

an

€

1

a−n = an

€

a

b

⎛

⎝ ⎜

⎞

⎠ ⎟−n

=b

a

⎛

⎝ ⎜

⎞

⎠ ⎟n

€

am

n = amn

OR

Evaluate:1.) 2.) 3.)

4.) 5.) 6.)

7.) 8.) 9.)

10.) 11.) 12.)

€

1253

€

1624

€

1691

2( )

0

€

2163( )

2

€

811

2 − 81−1

2

€

3433( )

−2

€

815

4

€

31

2 • 211

2

€

6251

4

€

2163

€

2−5 • 27

€

833

€

5

€

4

€

1

€

36

€

80

9

€

88

9

€

243

€

3 7

€

5

€

6

€

4

€

8

Express using rational exponents:1.) 2.) 3.)

4.) 5.) 6.)

7.) 8.) 9.)

10.) 11.) 12.)

€

a6b3

€

25a4b10

€

64s9t153

€

169x 5

€

15x 3y155

€

27x 4 y 34

€

x 5y 6

€

c 73

€

144x 6y10

€

27x10y 55

€

1024a3

€

36a8b54

€

a3b3

2

€

5a2b5

€

4s3t 5

€

13x5

2

€

151

5 x3

5 y 3

€

271

4 xy3

4

€

x5

2 y 3

€

c7

3

€

12x 3y 5

€

271

5 x 2y

€

32a3

2

€

61

2 a2b5

4

Express using radicals:1.) 2.) 3.)

4.) 5.) 6.)

7.) 8.) 9.)

10.) 11.) 12.)

€

x2

3

€

x4

7 y3

7

€

a1

6b2

3c1

2

€

15x1

3 y1

5

€

(5a)2

3 b5

3

€

3433( )

−2

€

641

3

€

y3

2

€

21

2 a3

2b5

2

€

x2

5 y3

5

€

s2t( )1

3v2

3

€

x 6y 3( )

12 z

32

€

x 23

€

x 4 y 37

€

ab4c 36

€

15 x 5y 315

€

b 25a2b23

€

149

€

4

€

y 3

€

ab2 2ab

€

x 2y 35

€

s2tv 23

€

x 3 y 3z3

Simplify:1.) 2.) 3.)

4.) 5.) 6.)

7.) 8.) 9.)

10.) 11.) 12.)

€

4x 2(4 x)−2

€

(y −2)4 • y 8

€

2x( )4

( )−2

€

a3b2 • a4b5

€

(5ac)1

3 (a2c 3)1

3

€

a4b83

€

x −2( )

3• x 3

( )−2

€

3y 3( ) 3y( )

3

€

x 7 • x 5 • x −7 • x −5

€

6a1

3( )

3

€

3x 2(3x)−2

€

2a( )1

3 a2b( )1

3

€

1

4

€

1

€

1

256x 8

€

a3b3 ab

€

ac 5c3

€

ab2 ab23

€

1

x12

€

81y 6

€

1

€

216a

€

13

€

a 2b3

General Form of Exponential Function y = b x where b > 1

Domain: All reals

Range:

y > 0 x-intercept:

None y-intercept:

(0, 1)

General Form of Exponential Function y = b (x + h) + k where b > 1

h moves graph left or right (opposite way)

k move graph up or down (expected way)

So y=3(x+2) + 3 moves the graph 2 units to the left and 3 units up

(0, 1) to (– 2, 4)

Graph:

€

y =1

3

⎛

⎝ ⎜

⎞

⎠ ⎟x

X Y

-2

-1

0

1

2

3

9

3

1

1/3

0.111

0.037

Decreasing for all of x

Graph:

€

y = 2x −1

X Y

-2

-1

0

1

2

3

0.125

0.25

0.5

1

2

4

Increasing for all of x

Graph:

€

y = −2x +1

X Y

-2

-1

0

1

2

3

-0.5

-1

-2

-4

-8

-16

Decreasing for all of x

Graph:

€

y = 2−x +1

X Y

-2

-1

0

1

2

3

8

4

2

1

0.5

0.25

Decreasing for all of x

Graph:

€

y > 2x

X Y

-2

-1

0

1

2

3

0.25

0.5

1

2

4

8

Graph:

€

y ≥ (0.5)x

X Y

-2

-1

0

1

2

3

4

2

1

0.5

0.25

0.125

Converting between Exponents & Logarithms

WRITE EACH EQUATION INTO A LOG FUNCTION:

WRITE EACH EQUATION INTO A EXPONENTIAL FUNCTION:

€

25 = 32

€

5−3 =1

125

€

6−3 =1

216

€

log2 32 = 5

€

log5

1

125= −3

€

log6

1

216= −3

€

log3 27 = 3

€

log4 16 = 2

€

log10

1

100= −2

€

33 = 27

€

42 =16

€

10−2 =1

100

€

log7 73

€

log10 0.001

€

3log3 6

Evaluate:

Evaluate:

Evaluate:

€

logb b−4

€

loga a

€

34 log3 4

Evaluate:

Evaluate:

Evaluate:

3

-3

6

-4

1

256

Properties of Logarithms (Shortcuts)

logb1 = 0 (because b0 = 1)

logbb = 1 (because b1 = b)

logb(br) = r (because br = br)

blog b M = M (because logbM = logbM)

Properties of Logarithms

logb(MN)= logbM + logbN

Ex: log4(15)= log45 + log43

logb(M/N)= logbM – logbN

Ex: log3(50/2)= log350 – log32

logbMr = r logbM

Ex: log7 (103) = 3 log7 10

If logbM = logbN Then M = N

log11 (1/8) = log11 8-1

TRY THESE TO SOLVE THESE:

€

log6 x = 2

€

log5

1

25= y

€

logx 64 = 3

€

log4 0.25 = x

€

log4 (2x −1) = log4 16

€

log10 10 = x€

62 = x

x = 36

€

5y =1

25

5y =1

52

5y = 5−2

y = −2

€

x 3 = 64

x = 4

€

4 x = 0.25

4 x =1

4

4 x = 4−1

x = −1

€

2x −1 =16

2x =17

x =17

2

€

10x = 10

10x =101

2

x =1

2

First I would change to exponential form!!

€

log3(4x −10) = log3(x −1)

€

log8(x 2 + 5x +14) = log8 0

It appears that we have 2 solutions here.If we take a closer look at the definition of a logarithm however, we will see that not only must we use positive bases, but also we see that the arguments must be positive as well. Therefore -2 is not a solution.

Our final concern then is to determine why logarithms like the one below are undefined.

Our final concern then is to determine why logarithms like the one below are undefined.

Can anyone give us an explanation ?

Can anyone give us an explanation ?2log ( 8)

One easy explanation is to simply rewrite this logarithm in exponential form. We’ll then see why a negative value is not permitted.

2log ( 8) y

€

2y = −8

What power of 2 would gives us -8 ?

Hence expressions of this type are undefined.

SOLVE EACH LOGARITHM EQUATION:

1.) 2.)

3.) 4.)

€

4 log8 x = log8 81

€

x = ±3 €

log5(y −12) + log5(y +12) = 2

€

y =13

€

log5 42 − log5 6 = log5 k

€

k = 7 €

log10 y =1

4log1016 +

1

2log10 49

€

y =14

€

y =13 or −13

Same Base

Solve: 4x-2 = 64x

4x-2 = (43)x

4x-2 = 43x

x–2 = 3x -2 = 2x -1 = x

If bM = bN, then M = N64 = 43

If the bases are already =, just solve the exponents

You Do

Solve 27x+3 = 9x-1

x 3 x 13 2

3x 9 2x 2

3 3

3 3

3x 9 2x 2

x 9 2

x 11

Review – Change Logs to Exponents

log3x = 2

logx16 = 2

log 1000 = x

32 = x, x = 9

x2 = 16, x = 4

10x = 1000, x = 3

Example

7xlog25 = 3xlog25 + ½ log225

log257x = log253x + log225 ½

log257x = log253x + log251

7x = 3x + 1 4x = 11

4x

Example 1 – Solving Simple Equations

a. 2x = 32 2x = 25 x = 5

b. ln x – ln 3 = 0 ln x = ln 3 x = 3

c. = 9 3– x = 32 x = –2

d. ex = 7 ln ex = ln 7 x = ln 7

e. ln x = –3 eln x = e– 3 x = e–3

f. log x = –1 10log x = 10–1 x = 10–1 =

g. log3 x = 4 3log3 x = 34 x = 81

Example 6 – Solving Logarithmic Equations

a. ln x = 2

eln x = e2

x = e2

b. log3(5x – 1) = log3(x + 7)

5x – 1 = x + 7

4x = 8

x = 2

Example 6 – Solving Logarithmic Equations

c. log6(3x + 14) – log6 5 = log6 2x

3x + 14 = 10x

–7x = –14

x = 2

cont’d

Given the original principal, the annual interest rate, and the amount of time for each investment, and the type of compounded interest, find the amount at the end of the investment.

1.) P = $1,250; r = 8.5%; t = 3 years; quarterly

2.) P = $2,575; r = 6.25%; t = 5 years, 3 months; continuously

€

A = P 1+r

n

⎛

⎝ ⎜

⎞

⎠ ⎟nt

€

A =1250 1+0.085

4

⎛

⎝ ⎜

⎞

⎠ ⎟4(3)

€

A = $1,608.77

€

A = Pert

€

A = 2575e0.0625(5.25)

€

A = $3,575.03