Exponential & Logarithmic Functions

REVIEW11.1 – 11.4

Properties of Exponents

am • an = am+ n

an = am −n

(am )n = am• n

(a • b)m = am • bm

ab ⎛ ⎝ ⎜

⎞ ⎠ ⎟m

a−n =1an

1a−n = an

ab ⎛ ⎝ ⎜

⎞ ⎠ ⎟−n

=ba ⎛ ⎝ ⎜

⎞ ⎠ ⎟n

n = amn

Evaluate:1.) 2.) 3.)

4.) 5.) 6.)

7.) 8.) 9.)

10.) 11.) 12.)

2163( )2

2 − 81−1

3433( )−2

2 • 211

2−5 • 27

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

4.) 5.) 6.)

7.) 8.) 9.)

10.) 11.) 12.)

25a4b10

64s9t153

169x 5

15x 3y155

27x 4 y 34

x 5y 6

144x 6y10

27x10y 55

1024a3

36a8b54

4s3t 5

12x 3y 5

5 x 2y

2 a2b5

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

4.) 5.) 6.)

7.) 8.) 9.)

10.) 11.) 12.)

3433( )−2

s2t( )1

x 6y 3( )1

x 4 y 37

ab4c 36

15 x 5y 315

b 25a2b23

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( )

a3b2 • a4b5

(5ac)1

3 (a2c 3)1

x −2( )3

• x 3( )−2

3y 3( ) 3y( )3

x 7 • x 5 • x−7 • x−5

3x 2(3x)−2

2a( )1

3 a2b( )1

1256x 8

a3b3 ab

ac 5c3

ab2 ab23

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 =13 ⎛ ⎝ ⎜

⎞ ⎠ ⎟x

Decreasing for all of x

Graph:

y = 2x −1

Increasing for all of x

Graph:

y = −2x +1

Graph:

y = 2−x +1

Graph:

y > 2x

Graph:

y ≥ (0.5)x

Converting between Exponents & LogarithmsWRITE EACH EQUATION INTO A LOG FUNCTION:

WRITE EACH EQUATION INTO A EXPONENTIAL FUNCTION:

25 = 32

5−3 =1

6−3 =1

log2 32 = 5

125= −3

216= −3

log3 27 = 3

log4 16 = 2

log101

100= −2

33 = 27

42 =16

10−2 =1

log7 73

log10 0.001

3log3 6

Evaluate:

logb b−4

loga a

34 log3 4

Evaluate:

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

log5125

logx 64 = 3

log4 0.25 = x

log4 (2x −1) = log4 16

log10 10 = x€

62 = xx = 36

5y =125

5y =152

5y = 5−2

y = −2

x 3 = 64x = 4

4 x = 0.25

4 x =14

4 x = 4−1

x = −1

2x −1 =162x =17

x =172

10x = 10

10x =101

First I would change to exponential form!!

log3(4 x −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.

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

log5 42 − log5 6 = log5 k

k = 7 €

log10 y =14

log1016 +12

log10 49

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 33 33x 9 2x 2x 9 2x 11

Review – Change Logs to Exponents

log3x = 2 logx16 = 2 log 1000 = x

32 = x, x = 9x2 = 16, x = 4

10x = 1000, x = 3

Example

7xlog25 = 3xlog25 + ½ log225 log257x = log253x + log225 ½

log257x = log253x + log251

7x = 3x + 1 4x = 11

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

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+rn

⎛ ⎝ ⎜

⎞ ⎠ ⎟nt

A =1250 1+0.085

4 ⎛ ⎝ ⎜

⎞ ⎠ ⎟4(3)

A = $1,608.77

A = Pert

A = 2575e0.0625(5.25)

A = $3,575.03

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