Skill 6 Exponential and

Logarithmic Functions

Skill 6a: Graphs of Exponential Functions Skill 6b: Solving Exponential Equations (not requiring logarithms) Skill 6c: Definition of Logarithms

Skill 6d: Graphs of Logarithms Skill 6e: Properties of logarithms (product rule, quotient rule, power rule, change of base) Skill 6f: Logarithmic Equations (not requiring exponentials) Skill 6g: Exponential and Logarithmic Equations requiring inverse operations

Skill 6a: Graphs of Exponential Functions An exponential function is defined as an expression with a constant base with a variable exponent. The following are examples of exponential functions:

𝑓(𝑥) = 2𝑥 𝑔(𝑥) = 𝜋3𝑥−1 ℎ(𝑥) = 3𝑥 + 2 In general an exponential function is of the form 𝑓(𝑥) = 𝑎𝑥, where 𝑎 > 0 and 𝑎 ≠ 1. Why is it necessary for 𝑎 > 0? What can't 𝑎 = 1? Say, 𝑎 = −1 then

𝑓 (1

2) = (−1)

12 = 𝑖, but we want real numbers.

http://mathforum.org/library/drmath/view/55604.html Say, 𝑎 = 1, then the range would be {1}.

1. Complete the table below for the function 𝑓(𝑥) = 2𝑥. Then graph the function at the right.

What is the domain of the basic exponential function? 𝐷 = (−∞, ∞)

What is the range of the basic exponential function? 𝑅 = (0, ∞)

What is the equation of the horizontal asymptote of the basic exponential function? 𝑦 = 0

2. Complete the table for the function 𝑓(𝑥) = 4𝑥. 3. Complete the table for 𝑓(𝑥) = (1

2)

𝑥.

Then graph the function above. Then graph the function above.

𝑥 𝑓(𝑥)

-2 4

-1 2

0 1

1 1

2

2 1

4

3 1

8

4 1

16

𝑥 𝑓(𝑥)

-2 1

4

-1 1

2

0 1

1 2

2 4

3 8

4 16

𝑥 𝑓(𝑥)

-2 1

16

-1 1

4

0 1

1 4

2 16

3 64

4 256

4. If 1 < 𝑎 < 𝑏. Sketch a graph that illustrates the difference between 𝑓(𝑥) = 𝑎𝑥 and 𝑔(𝑥) = 𝑏𝑥

5. If 0 < 𝑎 < 𝑏 < 1. Sketch a graph that illustrates the difference between 𝑓(𝑥) = 𝑎𝑥 and 𝑔(𝑥) = 𝑏𝑥

6. If 𝑎 > 1, how does the graph of 𝑓(𝑥) = 𝑎−𝑥 compare to the graphs of 𝑔(𝑥) = (1

𝑎)

𝑥 and ℎ(𝑥) =

1

𝑎𝑥 ?

They all have the same graph (they’re equivalent).

Match the function below with the correct graph.

A 7. 𝑦 = 3𝑥

C 8. 𝑦 = 3−𝑥

D 9. 𝑦 = 3𝑥 + 5

C 10. 𝑦 = (1

3)

𝑥

E 11. 𝑦 = (3

2)

𝑥

B 12. 𝑦 = (2

3)

𝑥

F 13. 𝑦 = 3𝑥−4

E 14. 𝑦 = (2

3)

−𝑥

C 15. 𝑦 = 1

3𝑥

A 16. 𝑦 = 1

3−𝑥

17. What is the domain, range, y-intercept, and the equation of the horizontal asymptote for 𝑓(𝑥) = 4𝑥+2 − 3.

Domain = (−∞, ∞)

Range = (−3, ∞)

The number 𝑒 is defined as the value of (1 +1

𝑛)

𝑛as n approaches infinity. 𝑒 is an irrational number, but to ten

decimal places it can be approximated as 2.7182818285. When 𝑒 is the base of an exponential function, it is called

the natural exponential function.

18. Sketch the graph of 𝑦 = 3𝑒0.5𝑥

Skill 6b: Solving Exponential Equations (not requiring logarithms) Some exponential equations can be solved by rewriting constants values in terms of the base. Solve for x: 1. 3𝑥 = 81 2. 63𝑥−7 + 4 = 40

3𝑥 = 34 𝑥 = 4

63𝑥−7 = 36 63𝑥−7 = 62 3𝑥 − 7 = 2 3𝑥 = 9 𝑥 = 3

3. 6 = 12(2𝑥) 4. 64𝑥 = 16

6

12= 2𝑥

2−1 = 2𝑥 𝑥 = −1

(43)𝑥 = 42 43𝑥 = 42 3𝑥 = 2

𝑥 =2

3

𝑥 𝑦

-3 ≈ .669

-2 ≈ 1.104

-1 ≈ 1.820

0 = 3

1 ≈ 4.946

2 ≈ 8.155

3 ≈ 13.445

5. 1252𝑥 = 25 6. 32𝑥+3 = 163𝑥−5

(53)2𝑥 = 52 56𝑥 = 52 6𝑥 = 2

𝑥 =1

3

(25)(𝑥+3) = (24)(3𝑥−5)

25𝑥+15 = 212𝑥−20 5𝑥 + 15 = 12𝑥 − 20 35 = 7𝑥 𝑥 = 5

Skill 6c: Definition of Logarithms A logarithm is defined as the inverse of an exponential function. 1. 𝑓(𝑥) = 2𝑥, A) What is 𝑓(3)? B) What is 𝑓−1(8)?

𝑓(3) = 23 = 8 𝑓−1(𝑥) = log2 𝑥 𝑓−1(8) = log2 8 = 3

The exponential equation 23 = 8 can be written as the logarithmic (or log) equation log2 8 = 3. Rewrite the following exponential equations as logarithmic equations.

2. 54 = 625 3. 3−5 = 1

243 4. 103 = 1000 5. 𝑒3 ≈ 20.086

log5 625 = 4 log31

243= −5 log 1000 = 3 ln 20.086 ≈ 3

Note that log10 𝑥 is usually written log 𝑥, so instead of writing log10 100 = 2, write log 100 = 2 . Also log𝑒 𝑥 is

written ln 𝑥.

Rewrite the following logarithmic equations as exponential equations.

5. log27 3 =1

3 6. log

1

10000= −4 7. log5 5 = 1 8. ln 1 = 0

2713 = 3 10−4 =

1

10000 51 = 5 𝑒0 = 1

Rewrite the following logarithmic equations as exponential equations and determine the value of x.

9. log4 𝑥 = 2 10. log4 64 = 𝑥 11. log2 𝑥 = −5

42 = 𝑥 𝑥 = 16

4𝑥 = 64

4𝑥 = 43 𝑥 = 3

2−5 = 𝑥

𝑥 =1

32

12. log25 5 = 𝑥 13. log𝑥 81 = 2 14. log 1,000,000 = 𝑥

25𝑥 = 5

52𝑥 = 51 2𝑥 = 1

𝑥 =1

2

𝑥2 = 81 𝑥 = 9 Note: The base must be greater than zero and not equal to one.

10𝑥 = 1,000,000

10𝑥 = 106 𝑥 = 6

Skill 6d: Graphs of Logarithms Since a logarithm is the inverse of an exponential function, the graph of a y = log2x is the reflection of the graph of

y = 2x across the line y = x.

𝑥 2𝑥 𝑥 log2 𝑥

−2 1

4

1

4 −2

−1 1

2

1

2 −1

0 1 1 0

1 2 2 1

2 4 4 2

For a basic logarithm:

Domain: (0, ∞)

Range: (−∞, ∞)

Vertical Asymptote: 𝑥 = 0

X - Intercept: 𝑥 = 1

State the domain, range, x-intercept, and give the equation of the vertical asymptote for each function below:

1. 𝑓(𝑥) = 5log2(𝑥) − 1 2. 𝑓(𝑥) = log5(𝑥 − 4)

Domain: (0, ∞) Range: (−∞, ∞) Domain: (4, ∞) Range: (−∞, ∞)

Vertical Asymptote: 𝑥 = 0 Vertical Asymptote: 𝑥 = 4

X - Intercept: 𝑥 = 215 X - Intercept: 𝑥 = 5

3. 𝑓(𝑥) = log3(9𝑥 − 17) 4. 𝑓(𝑥) = ln(1

2𝑥 + 2) − 4

Domain: (17

9, ∞) Range: (−∞, ∞) Domain: (−4, ∞) Range: (−∞, ∞)

Vertical Asymptote: 𝑥 =17

9 Vertical Asymptote: 𝑥 = −4

X - Intercept: 𝑥 = 2 X - Intercept: 𝑥 = 2𝑒4 − 4

Match the function below with the correct graph.

D 5. 𝑦 = log3 𝑥

B 6. 𝑦 = − log3 𝑥

C 7. 𝑦 = log3(−𝑥)

B 8. 𝑦 = log13

𝑥

F 9. 𝑦 = log3(𝑥 − 4)

E 10. 𝑦 = log32(𝑥)

A 11. 𝑦 = log3(𝑥

6)

B 12. 𝑦 = log3(1

𝑥)

Note:

A B C

D E F

Match the function below with the correct graph.

B 13. 𝑦 = ln 𝑥

C 14. 𝑦 = log5 𝑥

A 15. 𝑦 = log 𝑥

Skill 6e: Properties of Logarithms Derivation of the Product Rule log(𝑎𝑏) = 𝑦 , 𝑎 = 10𝑚 , and 𝑏 = 10𝑛 10𝑦 = 𝑎𝑏

10𝑦 = 10𝑚+𝑛 so, 𝑦 = 𝑚 + 𝑛 since 𝑎 = 10𝑚 and 𝑏 = 10𝑛, 𝑚 = log 𝑎 and 𝑛 = log 𝑏 So, log(𝑎𝑏) = 𝑦

log(𝑎𝑏) = 𝑚 + 𝑛 log(𝑎𝑏) = log 𝑎 + log 𝑏

Product Rule of Logarithms

log(𝑎𝑏) = log 𝑎 + log 𝑏

Also since log(𝑎𝑛) = log(𝑎 ∙ 𝑎 ∙ … ∙ 𝑎) = log 𝑎 + log 𝑎 + ⋯ log 𝑎 = 𝑛 log 𝑎

Power Rule of Logarithms

log(𝑎𝑛) = 𝑛 log 𝑎

And recall log1

𝑏= − log 𝑏

Quotient Rule of Logarithms

log (𝑎

𝑏) = log 𝑎 − log 𝑏

A C B

Rewrite the following using the properties of logarithms:

1. log2 32𝑥 2. log𝑥

100 3. log4 𝑥10

= log2 32 + log2 𝑥 = 5 + log2 𝑥

= log 𝑥 − log 100 = log 𝑥 − 2 = −2 + log 𝑥

= 10 log4 𝑥

4. log3𝑥3

𝑦 5. log5

1

𝑎𝑏 6. log7

√𝑥3

7

= log3 𝑥3 − log3 𝑦 = 3 log3 𝑥 − log3 𝑦

= log5(𝑎𝑏)−1 = − log5(𝑎𝑏) = −[log5 𝑎 + log5 𝑏] = − log5 𝑎 − log5 𝑏

= log7 √𝑥3

− log7 7

= log7(𝑥)13 − 1

= −1 −1

3log7 𝑥

Combine the following using the properties of logarithms into a single logarithm:

7. 4 log(𝑥) +log(𝑦)

2− log (𝑧) 8. 4 + log2 𝑥 9. 2 log3 𝑥 − 1 + log3 𝑦

= 4 log(𝑥) +1

2log(𝑦) − log (𝑧)

= log 𝑥4 + log 𝑦12 − log 𝑧

= log (𝑥4𝑦12) − log 𝑧

= log (𝑥4√𝑦

𝑧)

= log2 16 + log2 𝑥 = log2(16𝑥)

= log3 𝑥2 − log3 3 + log3 𝑦

= log3 (𝑥2

3) + log3 𝑦

= log3 (𝑦𝑥2

3)

If 𝐥𝐨𝐠𝟖 𝟓 ≈ 𝟎. 𝟕𝟕𝟒 and 𝐥𝐨𝐠𝟖 𝟑 = 𝟎. 𝟓𝟐𝟖, determine the following: 10. log8 25 11. log8 45 12. log8 320

= log8 52 = 2 log8 5 ≈ 2(0.774) = 1.548

= log8(5 ∙ 9) = log8 5 + log8 9 = log8 5 + log8 32 = log8 5 + 2 log8 3 ≈ 0.774 + 2(0.528) = 1.83

= log8(5 ∙ 64) = log8 5 + log8 64 = log8 5 + log8 64 = log8 5 + log8 82 = log8 5 + 2 log8 8 ≈ 0.774 + 2 = 2.774

13. log85

3 14. log8

125

8 15. log8 10

= log8 5 − log8 3 ≈ 0.774 − 0.528 = 0.246

= log8 125 − log8 8 = log8(53) − 1 = 3 log8(5) − 1 ≈ 3(0.774) − 1 = 1.322

= log8(2 ∙ 5) = log8 2 + log8 5

= log8 813 + log8 5

=1

3log8 8 + log8 5

≈1

3+ 0.774

≈ 1.107

Changing Bases: log𝑎 𝑏 = 𝑐 can be rewritten as 𝑎𝑐 = 𝑏 so, log 𝑎𝑐 = log 𝑏 or, 𝑐 log 𝑎 = log 𝑏

so, 𝑐 = log 𝑏

log 𝑎

log𝑎 𝑏 = 𝑐

log𝑎 𝑏 =log 𝑏

log 𝑎

So with just a 'log' or 'ln' button on a calculator, any logarthin can be found.

Change of Base Rule for Logarithms

log𝑎 𝑏 =log 𝑏

log 𝑎 or log𝑎 𝑏 =

ln 𝑏

ln 𝑎

Determine the following to four decimal places:

16. log4 60 17. log31

2 18. log7(−4)

=log 60

log 4

≈ 2.9534

=ln

12

ln 3

≈ −0.6309

= undefined

Skill 6f: Logarithmic Equations (not requiring inverse operations) Solve for x: 1. log(5) + log(𝑥) = log(3) + log (10) 2. log2 3 + log2 𝑥 = log2 5 + log2(𝑥 − 2)

log(5𝑥) = log (30) 5𝑥 = 30 𝑥 = 6

log2(3𝑥) = log2(5(𝑥 − 2))

log2(3𝑥) = log2(5𝑥 − 10) 3𝑥 = 5𝑥 − 10 −2𝑥 = −10 𝑥 = 5

3. log2(𝑥 − 4) = log2(5) − log2(𝑥) 4. 2 log 𝑥 = log 2 + log (3𝑥 − 4)

log2(𝑥 − 4) = log2 (5

𝑥)

𝑥 − 4 =5

𝑥

𝑥2 − 4𝑥 − 5 = 0 (𝑥 − 5)(𝑥 + 1) = 0 𝑥 = 5, 𝑥 = −1 is extraneous 𝑥 = 5

2 log 𝑥 = log(2(3𝑥 − 4)) log 𝑥2 = log(6𝑥 − 8)

𝑥2 = 6𝑥 − 8 𝑥2 − 6𝑥 + 8 = 0 (𝑥 − 2)(𝑥 − 4) = 0 𝑥 = 2, 𝑥 = 4

5. log3(5 − 2𝑥) = 3 6. log2(𝑥 + 2) + log2(𝑥) = 3

33 = 5 − 2𝑥 2𝑥 = −22 𝑥 = −11

log2(𝑥(𝑥 + 2)) = 3

log2(𝑥2 + 2𝑥) = 3 23 = 𝑥2 + 2𝑥 𝑥2 + 2𝑥 − 8 = 0 (𝑥 + 4)(𝑥 − 2) = 0 𝑥 = −4 (is extraneous) 𝑥 = 2 𝑥 = 2

Skill 6g: Logarithmic and Exponential Equations Exponential Functions and Logarithmic Functions are inverses of each other; 𝑓(𝑥) = 2𝑥 𝑓−1(𝑥) = log2 𝑥 𝑔(𝑥) = ln 𝑥 𝑔−1(𝑥) = 𝑒𝑥 Simplify the following expressions:

1. 3log3(4𝑥+3) 2. log6 6𝑥2 3. 𝑒ln(𝑥−5) 4. ln 𝑒(9−4𝑥)

= 4𝑥 + 3 = 𝑥2 = 𝑥 − 5 = 9 − 4𝑥

Solve each equation using inverse functions. Approximate solutions to 3 decimal places when needed.

5. 10𝑥 = 50 6. 4log(2𝑥) − 6 = 2 7. 𝑒5−𝑥 = 4

log 10𝑥 = log 50 𝑥 log 10 = log 50 𝑥 = log 50 𝑥 ≈ 1.699

log(2𝑥)4 − 6 = 2 log(24𝑥4) = 8 108 = 24𝑥4

𝑥4 =108

24

𝑥 = ±√108

24

4

𝑥 = ±102

2

𝑥 = −50 (is extraneous) 𝑥 = 50

ln 𝑒5−𝑥 = ln 4 (5 − 𝑥) ln 𝑒 = ln 4 5 − 𝑥 = ln 4 𝑥 = 5 − ln 4 𝑥 ≈ 3.614

8. 5ln(𝑥 − 2) = 15 9. 5𝑥+1 = 42𝑥+1 10. 101−𝑥 = 5𝑥

ln(𝑥 − 2) = 3

𝑒3 = 𝑥 − 2 𝑥 = 2 + 𝑒3 𝑥 ≈ 22.086

log 5(𝑥+1) = log 4(2𝑥+1) (𝑥 + 1) log 5 = (2𝑥 + 1) log 4

𝑥 log 5 + log 5 = 2𝑥 log 4 + log 4

𝑥 log 5 − 2𝑥 log 4 = log 4 − log 5 𝑥(log 5 − 2 log 4) = log 4 − log 5

𝑥 (log 5 − log 42) = log 4 − log 5

𝑥 log5

16= log

4

5

𝑥 =log(

4

5)

log(5

16)

𝑥 ≈ .192

log 10(1−𝑥) = log 5𝑥 (1 − 𝑥) log 10 = 𝑥 log 5 1 − 𝑥 = 𝑥 log 5 1 = 𝑥 + 𝑥 log 5 1 = 𝑥(1 + log 5)

𝑥 =1

(1+log 5)

𝑥 ≈ .589