1

Lesson 10.1: An Introduction to Logarithms

Learning Goals:

1) What is a logarithm?

2) How do we convert between logarithmic and exponential form?

Do Now: Find the inverse of each of the following: switch 𝑥 & y values!

1. 𝑓(𝑥) = 3𝑥 + 5 2. 𝑓(𝑥) = √𝑥 + 53

− 3 3. 𝑓(𝑥) = 2𝑥

𝑦 = 3𝑥 + 5 𝑦 = √𝑥 + 53

− 3 𝑦 = 2𝑥

𝑥 = 3𝑦 + 5 𝑥 = √𝑦 + 53 − 3 𝑥 = 2𝑦

Solve for 𝑦! Solve for 𝑦! Solve for 𝑦!

𝑥 − 5 = 3𝑦 𝑥 − 3 = √𝑦 + 53

𝑦 =𝑥−5

3= 𝑓−1(𝑥) (𝑥 − 3)3 = (√𝑦 + 5

3)3

(𝑥 − 3)3 = 𝑦 + 5

𝑦 = (𝑥 − 3)3 − 5 = 𝑓−1(𝑥)

What happens when you try to find the inverse of problem 3? Currently we don’t

know how to solve for 𝑦!

Defining Logarithmic Functions

The function 𝑦 = log𝑏𝑥 is the name we give the inverse of 𝑦 = 𝑏𝑥.

𝑦 = log𝑏𝑥 is the same as 𝑏𝑦 = 𝑥

A logarithm gives as its output (𝑦-value) the exponent we must raise 𝑏 to in order

to produce its input (𝑥-value).

How would you write the inverse to problem 3 using some form of the word

logarithm? 𝑥 = 2𝑦 so 𝑦 = log2𝑥

2

What are logarithms? Logarithms are another way to write an exponential

function!

log𝑏𝑁 = 𝑥 means 𝑏𝑥 = 𝑁, where 𝑏 > 0 and 𝑏 ≠ 1

Parts of a logarithm: base, anti-logarithm, and exponent

log 𝑏⏟base⏟

𝑥⏟anti−log

= 𝑦⏟exponent

Trick to remember is to “loop”:

log8𝑥 =1

3 log𝑥125 = −

3

2

81

3 = 𝑥 𝑥−3

2 = 125

𝑥 = 2 (𝑥−3

2)−2

3= (125)−

2

3

𝑥 =1

25

Exponential Form Logarithmic Form 𝑥

2𝑥 = 64 log264 = 𝑥 2𝑥 = 26 𝑥 = 6

𝑥4 = 81 log𝑥81 = 4 (𝑥4)

14 = (81)

14 𝑥 = ±3 so 𝑥 = +3

52 = 𝑥 log5𝑥 = 2 𝑥 = 25

72 = 𝑥 log7𝑥 = 2 𝑥 = 49

𝑥3 = 27 log𝑥27 = 3 (𝑥3)

13 = (27)

13 𝑥 = 3

3

Part 1: Write each exponential equation in logarithmic form.

1. 24 = 16 2. 81

3 = 2

log216 = 4 log82 =1

3

Part 2: Write each logarithmic equation in exponential form.

3. log432 =5

2 4. log𝑎

1

𝑎= −1

45

2 = 32 𝑎−1 =1

𝑎

Part 3: Solve each equation for 𝑥:

5. log2𝑥 = 4 6. log81𝑥 =1

2

24 = 𝑥 811

2 = 𝑥

𝑥 = 16 𝑥 = 9

7. log6𝑥 = −2 8. log64𝑥 =2

3

6−2 = 𝑥 642

3 = 𝑥

𝑥 =1

36 𝑥 = 16

Part 4: Evaluate each expression:

9. log25125 = 𝑥 set it = 𝑥 10. log51

25= 𝑥 set it = 𝑥

25𝑥 = 125 5𝑥 =1

25

(52)𝑥 = 53 5𝑥 = 25−1

2𝑥 = 3 5𝑥 = (52)−1

𝑥 =3

2 𝑥 = −2

4

11. log491

7= 𝑥 set it = 𝑥 12. log1000.1 = 𝑥 set it = 𝑥

49𝑥 =1

7 100𝑥 = 0.1

49𝑥 = 7−1 100𝑥 =1

10

(72)𝑥 = 7−1 100𝑥 = 10−1

2𝑥 = −1 (102)𝑥 = 10−1

𝑥 = −1

2 2𝑥 = −1

𝑥 = −1

2

Part 5: Solve each equation for 𝑏.

13. log𝑏27 = 3 14. log𝑏1

2= −1

𝑏3 = 27 𝑏−1 =1

2

√𝑏33

= √273

(𝑏−1)−1 = (1

2)−1

𝑏 = 3 𝑏 = 2

15. log𝑏√5 =1

4 16. log𝑏4 =

1

2

𝑏1

4 = √5 𝑏1

2 = 4

(𝑏1

4)4

= (51

2)4

(𝑏1

2)2

= (4)2

𝑏 = 52 = 25 𝑏 = 16

17. If 𝑓(𝑥) = log3𝑥, find 𝑓(1)

𝑓(1) means 𝑥 = 1!

5

𝑓(1) = log31 ← "logbase" in calc!

𝑦 = log31

3𝑦 = 1

𝑦 = 0

18. Verify the following by evaluating the logarithms:

log2(8)⏟ + log2(4)⏟ = log2(32)⏟ log2(8)⏟ + log2(4)⏟ = log2(32)⏟

2𝑥 = 8 2𝑥 = 4 2𝑥 = 32 3 + 2 = 5

2𝑥 = 23 2𝑥 = 22 2𝑥 = 25 5 = 5 √

𝑥 = 3 𝑥 = 2 𝑥 = 5

6

Homework 10.1: An Introduction to Logarithms

Write each of the following in logarithmic form.

1. 216 = 63 2. 4−2 = 0.0625 3. 6253

4 = 125

Write each of the following in exponential form.

4. 5 = log3243 5. −2 = log50.04 6. log49343 =3

2

Evaluate each of the following:

7. log6216 8. log232 9. 5 ∙ log88

Solve each equation for the variable.

10. 𝑎 = log416 11. log8𝑥 =1

2 12. log𝑏64 = 6

7

13. If 𝑓(𝑥) = log3𝑥, find 𝑓(81).

14. When $1 is invested at 6% interest, its value, 𝐴, after 𝑡 years is 𝐴 = 1.06𝑡.

Express 𝑡 in terms of 𝐴.

15. Can the value of log2(−4) be found? What about the value of log20? Why

or why not? What does this tell you about the domain of log𝑏𝑥?

16. Which of the following is equivalent to 𝑦 = log7𝑥?

(1) 𝑦 = 𝑥7 (2) 𝑥 = 𝑦7 (3) 𝑥 = 7𝑦 (4) 𝑦 = 𝑥1

7

17. Verify the following by evaluating the logarithms:

log4(4) + log4(16) = log4(64)

8

Lesson 10.2: Natural Logarithms

Learning Goals:

1) What is a logarithm?

2) What is the number 𝑒?

3) What is a natural logarithm?

4) What is a common logarithm?

5) How do we convert between logarithmic and exponential form?

Do Now: Which of the following are valid examples of logarithms?

log24 log14 log04 log327 log5(−25) log70

Valid Examples of Logarithms Invalid Examples of Logarithms

log327

log24

log04

log5(−25)

log14

log70

Definition of the Logarithm base 𝑏:

If three numbers, 𝐿, 𝑏, and 𝑥 with 0 < 𝑏 < 1 or 𝑏 > 1 are related by 𝑏𝐿 = 𝑥, then 𝐿

is the logarithm base 𝑏 of 𝑥, and we write log𝑏(𝑥) = 𝐿.

9

Compound Interest: You deposit $3500 in an account that pays 4% annual

interest compounded continuously. What is the balance after 1 year?

Solution: 𝑃 = 3500 𝑟 = .04 𝑡 = 1

𝑃 = amount you invest(principal) r = interest rate (%

100) 𝑡 = time

Use the formula for continuously compounded interest.

𝐴 = 𝑃 𝑒𝑟 𝑡 Write the formula

= 3500 𝑒(.04∙1) Substitute for 𝑃, 𝑟, and 𝑡

𝐴 = 3642.84 The balance at the end of 1 year is $3642.84

Complete the following example:

You deposit $4800 in an account that pays 6.5% annual interest compounded

continuously. What is the balance after 3 years?

When it says compounded continuously use base of 𝑒!

Solution: 𝑃 = 4800 𝑟 = .065 𝑡 = 3

𝑃 = amount you invest(principal) r = interest rate (%

100) 𝑡 = time

Use the formula for continuously compounded interest.

The Natural Base 𝒆

The natural base 𝑒 is irrational. It is defined as follows:

As 𝑛 approaches +∞, (1 +1

𝑛)𝑛

approaches

𝑒 ≈ 2.718. 2nd → ÷ → 𝑒

Natural Base Functions

A function of the form 𝑦 = 𝑎 𝑒𝑟 𝑥 is called a natural base exponential function.

If 𝑎 > 0 and 𝑟 > 0, the function is an exponential GROWTH function.

If 𝑎 > 0 and 𝑟 < 0, the function is an exponential DECAY function.

10

𝐴 = 𝑃 𝑒𝑟 𝑡 Write the formula

= 4800 𝑒(.065∙3) Substitute for 𝑃, 𝑟, and 𝑡

𝐴 = 5833.49 The balance at the end of 3 years is $5833.49

The natural logarithms, gives an exponent as its output. In fact, it gives the

power that we must raise 𝑒 to in order to get the input.

𝑦 = ln𝑥 ln𝑥 = log𝑒𝑥 ln𝑥 ≠ ln𝑒𝑥

Without using your calculator, determine the values of each of the following.

1. ln(𝑒) = 𝑥 2. ln1 = 𝑥

log𝑒𝑒 = 𝑥 log𝑒1 = 𝑥

𝑒𝑥 = 𝑒1 𝑒𝑥 = 1

𝑥 = 1 𝑥 = 0

3. ln(𝑒5) = 𝑥 4. ln√𝑒 = 𝑥

log𝑒𝑒5 = 𝑥 log𝑒√𝑒 = 𝑥

𝑒𝑥 = 𝑒5 𝑒𝑥 = √𝑒

𝑥 = 5 𝑥 =1

2

Rewrite each of the following in logarithmic form.

5.. 𝑒5 = 𝑥 6. 𝑒𝑥 = 15

log𝑒𝑥 = 5 log𝑒15 = 𝑥

ln𝑥 = 5 ln15 = 𝑥

The Natural Logarithm

The inverse of 𝑦 = 𝑒𝑥: 𝑦 = ln𝑥 𝑦 = loge𝑥

11

Common and Natural Logarithms {Calculator Information}

Common Logarithm: A logarithm with a base of 10.

On the graphing calculator, the Log key is used to display the common logarithm

of a number.

Examples: Find the common log of each number to four decimal places.

1. 279 = log(279) = 2.4456 2. 0.0005 = log(. 0005) = −3.3010

When I find the common log of a number, it means that

102.4456 ≈ 279 and 10−3.3010 ≈ 0.0005

How can we find the number if we know the common log of the number?

log(𝑥) = log10(𝑥)

Examples: Find to the nearest thousandth each number 𝑁 whose common

logarithm is given.

3. log𝑁 = 3.9294 4. log𝑁 = −1.7799

log10𝑁 = 3.9294 log10𝑁 = −1.7799

103.9294 = 𝑁 10−1.779 = 𝑁

12

On the graphing calculator, the ln key is used to display the natural logarithm of a

number.

Find the natural log of each number to four decimal places.

5. ln52 = 3.9512 6. 2ln16 = 5.5452

When I find the natural log of a number, it means that 𝑒3.9512 ≈ 52

How can we find the number if we know the natural log of the number?

ln(𝑥) = log𝑒(𝑥)

Examples: Find to four decimal places the antilogarithm of the given logarithm.

7. ln𝑥 = −.5373 8. ln𝑥 = .8297

𝑒−.5373 = 𝑥 𝑒 .8297 = 𝑥

𝑥 = .5843 𝑥 = 2.2926

Compute the value of each logarithm. Verify your answers using an exponential

statement.

1. log13(13) = 1 2. log√7 (1

49) = −4

13𝑥 = 13 √7𝑥=

1

49

13𝑥 = 131 (71

2)𝑥

= 49−1

𝑥 = 1 (71

2)𝑥

= (72)−1

1

2𝑥 = −2

𝑥 = −4

13

What is wrong with the following questions?

3. log2(0) 4. log3 (−1

3)

They can’t be evaluated!

Find the value of each of the following.

5. If 𝑥 = log2(8) and 𝑦 = 2𝑥, 6. If 𝑟 = 26 and 𝑠 = log2(𝑟),

find the value of 𝑦. find the value of 𝑠.

2𝑥 = 8 and 𝑦 = 2𝑥, then 𝑦 = 8 2𝑠 = 𝑟 sub in the value of 𝑟!

2𝑠 = 26

𝑠 = 6

Rewrite each of the following in exponential form.

7. ln25 = 𝑥 8. ln𝑥 = 52 9. log45 = 𝑥 10. log𝑥 = .32

𝑒𝑥 = 25 𝑒52 = 𝑥 10𝑥 = 45 10𝑥 = .32

Find 𝑥 using your calculator. Round all answers to the nearest hundredth.

11. ln45 = 𝑥 12. log75 = 𝑥 13. ln𝑥 = 3.2958 14. log𝑥 = −.4881

𝑥 = 3.81 𝑥 = 1.88 𝑥 = 27.00 𝑥 = 0.33

15. Solve for 𝑥: 3 log(𝑥 + 4) = 6 16. Solve for 𝑥 to the nearest

3 log(𝑥+4)

3=6

3 thousandth:

log(𝑥 + 4) = 2 4 + ln(𝑥 − 1) = 4.8200

102 = 𝑥 + 4 ln(𝑥 − 1) = .82

100 = 𝑥 + 4 𝑒 .82 = 𝑥 − 1

𝑥 = 96 𝑥 = 𝑒 .82 + 1

You can always check your answers! 𝑥 = 3.270

14

Homework 10.2: Natural Logarithms

1. When Kyle was born, his grandparents invested $5000 in a college fund that

paid 4% compounded continuously. What is the value of the account after 18

years? {𝐴 = 𝑃𝑒𝑟 𝑡}

2. A hot liquid is cooling in a room whose temperature is constant. Its

temperature can be modeled using the exponential function shown below. The

temperature, 𝑇, is in degrees Fahrenheit and it is a function of the number of

minutes, 𝑚, it has been cooling.

𝑇(𝑚) = 101 𝑒−0.03 𝑚 + 67

a) What was the initial temperature of the water at 𝑚 = 0?

b) How do you interpret the statement that 𝑇(60) = 83.7?

15

Determine the value for each of the following logarithms.

3. log2 (1

64) 4. log48 5. log2√4

5

6. ln(𝑒4) 7. ln√𝑒3

8. ln𝑒

Rewrite in logarithmic form.

9. 𝑒7 = 𝑥 10. 𝑒𝑥 = 7 11. 2𝑥 = 5

12. log927 13. log𝑥(𝑥3)

Find 𝑥 in each of the following. Round answers to the nearest hundredth.

14. ln𝑥 = 3.3534 15. log𝑥 = 1.7218

16. log0.528 = 𝑥 17. ln51.3 = 𝑥

16

Lesson 10.3: Properties of logarithms

Learning Goal: How do we use the properties to expand and simplify log

expressions?

Do Now: Answer the following questions in order to prepare for today’s lesson.

1. Simplify: (3𝑥5𝑦4)

2∙(2𝑠4𝑦)

6𝑥5𝑦6=(32∙𝑥10∙𝑦8)(2𝑥4𝑦)

6𝑥5𝑦6=18𝑥14𝑦9

6𝑥5𝑦6= 3𝑥9𝑦3

(1) Power Law (2) Product Law (3) Quotient Law

2. Write each expression in radical form.

a. 71

2 index = 2 b. 25

3 index = 3 c. (3𝑚)7

4 index = 4

√7 √253

√(3𝑚)74

3. Write each expression in exponential form.

a. √23

= 21

3 b. (√10)3= 10

3

2 c. √𝑥5𝑦4

= 𝑥5

4𝑦1

4

4. What are logarithms? Logs are exponents!

5. Label the base, exponent, and anti-logarithm: logb𝑁 = 𝑥

17

Using the properties of exponents, we can arrive at the properties of

logarithms.

Properties of Exponents Properties of Logarithms

Multiplication Law

𝑏𝑛 ∙ 𝑏𝑚 = 𝑏𝑛+𝑚 logb(𝑚 ∙ 𝑛) = logb𝑚+ logb𝑛

Division Law 𝑏𝑛

𝑏𝑚= 𝑏𝑛−𝑚 logb (

𝑚

𝑛) = logb𝑚− logb𝑛

Power Law

(𝑏𝑛)𝑚 = 𝑏𝑛∙𝑚 log𝑏𝑚𝑟 = 𝑟logb𝑚

When separating logs, the power law is done last!

Directions: Rewrite each expression as sums and differences in terms of

log(𝑥) , log(𝑦) , and log (𝑧)

1. log𝑥3

𝑦= 3log𝑥 − log𝑦

2. log𝑏𝑥√𝑦 = log𝑏𝑥 + log𝑏√𝑦 = log𝑏𝑥 + log𝑏𝑦1

2 = log𝑏𝑥 +1

2log𝑏𝑦

3. log𝑥

𝑦𝑧= log𝑥 − (log𝑦 + log𝑧) = log𝑥 − log𝑦 − log𝑧

4. log(√𝑥7𝑦3) = log(√7 ∙ √𝑦3) = log√7 + log√𝑦3 = log71

2 + log𝑦3

2

=1

2log7 +

3

2log𝑦 or

1

2(7log𝑥 + 3log𝑦)

5. log√𝑥

𝑦63

= log𝑥1

3 − log𝑦2 =1

3log𝑥 − 2log𝑦

6. log (√𝑥𝑦73

𝑥2𝑧) = log (𝑥

1

3) + log𝑦7

3 − log𝑥2 − log𝑧 =

1

3log𝑥 +

7

3log𝑦 − 2log𝑥 − log𝑧

18

Directions: Use the properties of logarithms to rewrite each expression in an

equivalent form containing a single logarithm.

1. 2log5𝑚+ log5𝑛 = log5𝑚2 + log5𝑛 = log5(𝑚

2𝑛)

2. 1

2log𝑚 − 3log𝑛 = log√𝑚 − log𝑛3 = log (

√𝑚

𝑛3)

3. log2 + log𝐼 − log𝑇 = log (2𝐼

𝑇)

4. 2log𝑥 − (3log𝑦 + log𝑧) = log𝑥2 − (log𝑦3 + log𝑧) = log (𝑥2

𝑦3𝑧)

5. 2log𝑥 + 3log𝑦 −1

2log𝑧 = log𝑥2 + log𝑦3 − log√𝑧 = log (

𝑥2𝑦3

√𝑧)

6. 1

3(log(𝑥) − 3 log(𝑦) + log (𝑧)) = √log (

𝑥

𝑦3𝑧)

3

When proving equations, you must get the left side = to the right side!

Directions: Use properties of logarithms to show that:

1. log(26) − log(13) = log (2) 2. log(3) + log(4) + log(5) − log(6) = 1

log (26

13) = log2 log (

3(4)(5)

6) = 1

log2 = log2 √ log10 = 1

1 = 1 √

3. 1

2log(25) + log(4) = log (20) 4. log (

1

2−1

3) + log(2) = log (

1

3)

log√25 + log4 = log20 log (1

6) + log2 = log (

1

3)

log5 + log4 = log20 log (1

6∙ 2) = log (

1

3)

log(5 ∙ 4) = log 20 log (1

3) = log (

1

3) √

log20 = log20 √

19

5. log (1

3−1

4) + (log (

1

3) − log (

1

4)) = log (

1

9)

log (1

12) + log (

1

31

4

) = log (1

9)

log (1

12) + log (

4

3) = log (

1

9)

log (1

12∙4

3) = log (

1

9)

log (1

9) = log (

1

9) √

20

Homework 10.3: Properties of logarithms

1. The expression log4𝑥 is equivalent to:

(1) log𝑥4 (2) 4log𝑥 (3) log4 + log𝑥 (4) (log4)(log𝑥)

2. log √𝑥𝑦

𝑧 is equal to:

(1) 1

2log𝑥 +

1

2log𝑦 − log𝑧 (2)

1

2log𝑥 + log𝑦 − log𝑧

(3) 1

2(log𝑥 + log𝑦 − log𝑧) (4)

1

2log𝑥𝑦

log𝑧

3. The expression 3log𝑥 −1

2log𝑦 is equal to

(1) log𝑥3

𝑦2 (2) log

𝑥3

√𝑦 (3) log√

3𝑥

𝑦 (4)

log3𝑥1

2log𝑦

4. Apply properties of lograrithms to rewrite the following expressions as a single

logarithm or number.

a. 1

2ln(36) + ln (2)

b. ln(4) − 3ln (1

3) + ln (2)

c. ln(5) +3

5ln(32) − ln (4)

5. Apply properties of logarithms to rewrite each expression as a sum of terms

involving numbers, log(𝑥) , log(𝑦) , and log (𝑧), where 𝑥, 𝑦, and 𝑧 are positive real

numbers.

a. log (𝑥𝑦𝑧3) b. log (√𝑥3

𝑦𝑧

3) c. log(3𝑥4√𝑦5)

6. Use properties of logarithms to show that:

a. log(2) − log (1

13) = log (26) b.

1

2log(16) + log(3) + log (

1

4) = log (3)

21

Lesson 10.4: More Properties of Logarithms and Substitution

Learning Goals:

1) How do we use the properties to expand and simplify log expressions?

2) How do we simplify a logarithmic expression using powers of 10 rules?

Do Now: Answer the following questions in order to prepare for today’s lesson.

1. What is a common logarithm? How do you know when you have a common

logarithm? A common log has no base visible so we know base = 10 for

example log𝑥 = log10𝑥

2. Complete the following table of logarithms by using a calculator; then, answer

the questions that follow.

𝑥 log (𝑥) 1,000,000 106 = 6

100,000 105 = 5

10,000 104 = 4

1,000 103 = 3

100 102 = 2

10 101 = 1

a) What is log (1)? How does that follow from the definition of a base-10

logarithm?

log(1) = 0 100 = 1 count # of zeros since the base = 10

b) What is log(10𝑘) for an integer 𝑘? How does that follow from the definition of

a base- 10 logarithm?

Write a conjecture using logarithmic notation.

log(10𝑘) = 𝑘 𝑘 is the # of zeros!

22

Logarithm Rules from Previous Lesson

Multiplication Law log𝑏(𝑚𝑛) = log𝑏𝑚+ logb𝑛

Division Law

log𝑏 (𝑚

𝑛) = log𝑏𝑚 − logb𝑛

Power Law log𝑏𝑚

𝑟 = 𝑟log𝑏𝑚

New Logarithm Rules

When expanding logarithms, you must be aware of special logs that can be

reduced!

Below are two special cases:

log(10𝑘) = 𝑘

ln(𝑒) = 1

𝑘 is the # of zeros or exponent of 10. ln(𝑒) = log𝑒𝑒 = 1 so 𝑒1 = 𝑒

Directions: Apply properties of logarithms to rewrite each expression as a sum of

terms involving numbers, log(𝑥) , log(𝑦) , and log (𝑧), where 𝑥, 𝑦, and 𝑧 are positive

real numbers.

a. log(1000√𝑥) = log1000 + log√𝑥 = 3 +1

2log𝑥

b. log (100𝑥2

𝑦3) = log100 + 2log𝑥 − 3log𝑦 = 2 + 2log𝑥 − 3log𝑦

c. log (1

10𝑥2𝑧) = log1 − (log10 + 2log𝑥 + log𝑧) = 0 − (1 + 2log𝑥 + log𝑧)

= −1 − 2log𝑥 − log𝑧

Directions: Write each expression as a sum or difference of constants and

logarithms of simpler terms. Remember: ln𝑒 = 1

a. ln(3𝑒) = ln3 + ln𝑒 = ln3 + 1

b. ln (𝑒4

𝑥𝑦) = ln𝑒4 − (ln𝑥 + ln𝑦) = 4ln𝑒 − ln𝑥 − ln𝑦 = 4 − ln𝑥 − ln𝑦

c. ln (√5𝑥3

𝑒2) = ln(√5𝑥3) − ln𝑒2 = ln (5

1

2 ∙ 𝑥3

2) − ln𝑒2 =

1

2ln5 +

3

2ln𝑥 − 2ln𝑒 =

1

2ln5 +

3

2ln𝑥 − 2 or

1

2(ln5 + 3ln𝑥) − 2

23

Substitution with Logarithms

When asked to substitute using logarithms, you must EXPAND the

logarithms first by using the properties of logarithms.

Example 1: If log2 = 𝑥 and log3 = 𝑦, express log6 in terms of 𝑥 and 𝑦.

log6 = log(2 ∙ 3) = log2 + log3 = 𝑥 + 𝑦

Example 2: If ln2 = 𝑥 and ln3 = 𝑦, express ln4

9 in terms of 𝑥 and 𝑦.

ln4 − ln9 = ln22 − ln32 = 2ln2 − 2ln3 = 2𝑥 − 2𝑦

Example 3: If log5 = 𝑎, then log250 can be expressed as:

(1) 50𝑎 (2) 2𝑎 + 1 (3) 10 + 2𝑎 (4) 25𝑎

log250 = log(10 ∙ 25) = log10 + log25 = 1 + log52 = 1 + 2log5 = 1 + 2𝑎

Example 4: If log𝑥 = 𝑎, log𝑦 = 𝑏, and log𝑧 = 𝑐, rewrite log𝑥2𝑦

√𝑧 in terms of

𝑎, 𝑏, and 𝑐.

log𝑥2 + log𝑦 − log√𝑧 = 2log𝑥+ log𝑦 −1

2log𝑧 = 2𝑎 + 𝑏 −

1

2𝑐

Practice

1. Use the approximate logarithm values below to estimate the value of each of

the following logarithms.

log(2) = 0.3010

log(3) = 0.4771

log(7) = 0.8451

a. log12 = log(4 ∙ 3) = log(22 ∙ 3) = 2log2 + log3 = 2(. 3010) + .4771 = 1.0791

b. log (1

9) = log1 − log9 = 0 − log32 = −2log3 = −2(. 4771) = −.9542

c. log(70) = log(7 ∙ 10) = log7 + log10 = .8451 + 1 = 1.8451

24

2. Given: log𝑏3 = 𝑝 and log𝑏5 = 𝑞

Express each of the following in terms of 𝑝 and 𝑞.

a. log𝑏45 = log𝑏(9 ∙ 5) = log𝑏(32) + log𝑏5 = 2log𝑏3 + log𝑏5 = 2𝑝 + 𝑞

b. log𝑏√3

5= log

𝑏

√3

√5= log

𝑏√3 − log𝑏√5 =1

2log

𝑏3 −

1

2log

𝑏5 =

1

2𝑝 −

1

2𝑞 or

1

2(𝑝 − 𝑞)

c. log𝑏27

√53 = log𝑏27 − log𝑏√5

3= log𝑏(3

3) − log𝑏 (51

3) =

3log𝑏3 −1

3log𝑏5 = 3𝑝 −

1

3𝑞

d. log𝑏9

√53 = log𝑏9 − log𝑏√5

3= log𝑏(3

2) − log𝑏 (51

3) =

2log𝑏3 −1

3log𝑏5 = 2𝑝 −

1

3𝑞

3. If ln𝑎 = 2 and ln𝑏 = 3, what is the numerical value of ln (𝑒√𝑎

𝑏3)?

ln𝑒 +1

2ln𝑎 − 3ln𝑏 = 1 +

1

2(2) − 3(3) = 1 + 1 − 9 = −7

25

Homework 10.4: More Properties of Logarithms and Substitution

1. If log𝑎 = 𝑥 and log𝑏 = 𝑦, then log√𝑎𝑏 is equivalent to:

(1) 1

2𝑥 + 𝑦 (2)

1

2(𝑥 + 𝑦) (3)

1

2𝑥𝑦 (4)

1

4𝑥𝑦

2. Given: ln2 = 𝑥 and ln3 = 𝑦

Express each of the following in terms of 𝑥 and 𝑦: ln √2

9

3. Express each of the following as a single logarithm.

(a) log(8) + log (1

4)

(b) 1

3log(8) + 2 log(3) − log (6)

(c) 3 log(𝑥) − (2 log(𝑦) +1

2log (𝑧))

4. In the following expression, 𝑥, 𝑦, and 𝑧 represent positive real numbers. Use

properties of logarithms to rewrite the expression in an equivalent form

containing only log(𝑥) , log(𝑦) , log(𝑧), and numbers.

log√(𝑥3𝑦2

10𝑧)

5. Rewrite the expression as sums and differences in terms of

ln(𝑥) , ln(𝑦) , and ln (𝑧).

ln (𝑒

𝑦𝑧3)

26

Lesson 10.5: Solving Exponential Equations Using Logarithms

Learning Goal: How do we use logarithms to solve an exponential equation?

a) Solve the following equation for 𝑥: 43𝑥−1 = 32

Get like bases! (22)3𝑥−1 = 25

2(3𝑥 − 1) = 5

6𝑥 − 2 = 5

6𝑥 = 7

You can always check your answer! 𝑥 =7

6

b) What happens when you try to solve 52𝑥 = 20 for 𝑥?

You cannot get like bases, so we will need to use logs!

How can we solve exponential equations when you cannot get like bases?

1) Look to isolate the base with the variable exponent.

2) Take the common log of both sides

3) Use the power law to move the exponent

4) Solve for the variable

1. 2 ∙ 52𝑥 = 20

2∙52𝑥

2=20

2

52𝑥 = 10

2𝑥log5 = log10

2𝑥log5

log5=log10

log5

2𝑥 = 1.430676558

𝑥 = .715338279

27

Solve each equation for the given variable and express your answer to the

nearest hundredth.

1. 72𝑤 + 8 = 34 2. 𝑒2𝑥 = 5

72𝑤 = 26 2𝑥ln𝑒 = ln5 or 2𝑥log𝑒 = log5

2𝑤log7 = log26 2𝑥 = ln5 or 2𝑥 =log5

log𝑒

2𝑤log7

log7=log26

log7 𝑥 = .80

2𝑤 = 1.67433041

𝑤 = 0.84

3. 52+𝑥 − 5𝑥 = 10 4. 32𝑥−3 = 2𝑥+4

GCF: 5𝑥 (2𝑥 − 3)log3 = (𝑥 + 4)log2

5𝑥(52 − 1) = 10 2𝑥 − 3 = (𝑥 + 4)log2

log3

5𝑥(25 − 1) = 10 2𝑥 − 3 = (𝑥 + 4)(.6309297536)

5𝑥(24) = 10 2𝑥 − 3 = .63𝑥 + 2.52

5𝑥 =5

12 1.37𝑥 = 5.52

𝑥log5 = log5

12 𝑥 = 4.03

𝑥 = −0.54

Solve each equation for the given variable and express your answer to the

nearest hundredth.

5. 5𝑒3𝑥+4 = 15 6. 5𝑥+3 = 42𝑥−1

𝑒3𝑥+4 = 3 (𝑥 + 3)log5 = (2𝑥 − 1)log4

(3𝑥 + 4)ln𝑒 = ln3 𝑥 + 3 = (2𝑥 − 1)log4

log5

3𝑥 + 4 = ln3 (𝑥 + 3) = (2𝑥 − 1)(.8613531161)

28

𝑥 =ln (3)−4

3 𝑥 + 3 = 1.72𝑥 − .86

𝑥 = −.9671292371 −.72𝑥 = −3.86

𝑥 = −0.97 𝑥 = 5.36

7. 2(4𝑥) + 4𝑥+1 = 342 8. √3 ∙ 33𝑥 = 9

GCF: 4𝑥 31

2 ∙ 33𝑥 = 32

4𝑥(2 + 41) = 342 31

2+3𝑥 = 32

4𝑥(6) = 342 1

2+ 3𝑥 = 2

4𝑥 = 57 3𝑥 =3

2

𝑥log4 = log57 𝑥 =1

2

𝑥 = 2.92

Applications of Logarithms

9. You are investing $600 at an interest rate of 7.5% compounded continuously.

Using the formula, 𝐴 = 𝑃𝑒𝑟 𝑡, where 𝐴 is your final amount, 𝑃, is your starting

amount, 𝑟 is the interest rate, and 𝑡 is the time in years, determine how long it will

take you to reach $1000, to the nearest tenth of a year?

1000 = 600 𝑒 .075∙𝑡 50 = 80(0.98)𝑡,

5

3= 𝑒 .075∙𝑡

5

8= (0.98)𝑡

ln (5

3) = .075𝑡(ln𝑒) log (

5

8) = 𝑡 log(0.98)

𝑡 = 6.8 𝑡 = 23.26438837

1998 + 23 = 2021

10. A population of wolves in a county is represented by the equation

𝑃(𝑡) = 80(0.98)𝑡, where 𝑡 is the number of years since 1998. In what year will

50 wolves be reached? See above to right for answer

29

Homework 10.5: Solving Exponential Equations Using Logarithms

1. Solve the following exponential equation and express your answer to the

nearest hundredth: (1

2)

𝑥

3+1=1

6.

2. Solve the following equation for 𝑥 and express your answer to three decimal

places:

2𝑥+1 = 31−𝑥

3. Find the solution to the exponential equation below to the nearest hundredth:

4(2)𝑥 − 3 = 17

30

4. The logarithmic expression ln (√𝑒

𝑦3) can be rewritten:

(1) 3ln𝑦 − 2 (2) 1−6ln𝑦

2 (3)

ln𝑦−6

2 (4) √ln𝑦 − 3

5. The savings bank account can be modeled using 𝑆(𝑡) = 1250 𝑒0.045 𝑡, where 𝑡 is the number of years the money has been in the account. Determine, to the

nearest tenth of a year, how long it will take for the amount of savings to double

from the initial amount deposited of $1250.

6. Matt places $1200 in an investment account earning an annual rate of 6.5%,

compounded continuously. Using the formula 𝐴 = 𝑃𝑒𝑟 𝑡, how long will it take for

his initial investment to quadruple in value? Round to the nearest year.

31

Lesson 10.6: Solving Logarithmic Equations

Learning Goals:

1) How do we solve logarithmic equations?

2) How do we check for extraneous roots?

Warm Up:

Using what we have learned about solving for an exponent, answer each of the

following questions:

a) Solve the following equation for 𝑡: 𝑎 ∙ 𝑏𝑐 𝑡 = 𝑑

𝑎∙𝑏𝑐 𝑡

𝑎=𝑑

𝑎

𝑏𝑐 𝑡 =𝑑

𝑎

𝑐𝑡(log𝑏) = log (𝑑

𝑎)

𝑐 𝑡 (log𝑏)

𝑐(log𝑏)=

log(𝑑

𝑎)

𝑐(log𝑏)

𝑡 =log(

𝑑

𝑎)

𝑐(log𝑏)

RECALL

𝑏𝐿 = 𝑥 can be written in the form log𝑏(𝑥) = 𝐿 where

0 < 𝑏 < 1 or 𝑏 > 1 and 𝑥 > 0

*This means the base cannot be 0, 1, or a negative number and the

anti-logarithm cannot be 0 or a negative number.

32

b) Solve the following equation for 𝑎: 3 + 𝑥𝑎+𝑏 = 𝑐

𝑥𝑎+𝑏 = 𝑐 − 3

(𝑎 + 𝑏)log𝑥 = log (𝑐 − 3)

𝑎 + 𝑏 =log (𝑐−3)

log𝑥

𝑎 =log (𝑐−3)

log𝑥− 𝑏

There are 3 types of log equations:

Type Procedure Example

1 log “loop” Cannot loop until you isolate!

log((2𝑥 + 5)2) = 4

104 = (2𝑥 + 5)2 3 − log2(𝑥 − 1) = 0 −log2(𝑥 − 1) = −3 log2(𝑥 − 1) = 3

23 = 𝑥 − 1

1

2log(𝑥 + 2) = 2

log(𝑥 + 2)12 = 2

log(𝑥 + 2)12∙2 = 22

log(𝑥 + 2) = 4

104 = 𝑥 + 2

1 log on each side

“cancel” Logs must have like bases!

2 ln(𝑥 + 2) = ln (−𝑥) ln(𝑥 + 2)2 = ln (−𝑥) (𝑥 + 2)2 = −𝑥

2 or more logs on one side

“combine” log𝑚 + log𝑛 = log (𝑚 ∙ 𝑛)

log𝑚 − log𝑛 = log (𝑚

𝑛)

log2(3𝑥) + log2(4) = 4 log2(12𝑥) = 4 24 = 12𝑥

2log5𝑥 − log55 = log5125

log5𝑥2 − log5 = log5125

𝑥2

5= 125

33

Practice! Determine which of the three methods would be most appropriate

before solving each problem. Be sure to check for extraneous solutions.

1. log2(9𝑥2 + 30𝑥 + 25) = 8 2. ln(𝑥 + 2) = ln(12) − ln(𝑥 + 3)

28 = 9𝑥2 + 30𝑥 + 25 𝑥 + 2 =12

𝑥+3 Cross Multiply!

256 = 9𝑥2 + 30𝑥 + 25 𝑥(𝑥 + 2)(𝑥 + 3) = 12

0 = 9𝑥2 + 30𝑥 − 231 𝑥2 + 5𝑥 + 6 = 12

0 = 3(3𝑥2 + 10𝑥 − 77) 𝑥2 + 5𝑥 − 6 = 0

Quadratic Formula or AC Method (𝑥 − 1)(𝑥 + 6) = 0

𝑥 =11

3 & − 7 √ 𝑥 = 1 & − 6

Always check! Omit −6 so the only answer is 𝑥 = 1

𝑥 + 2 > 0 𝑥 + 3 > 0

−6 + 2 > 0 −6 + 3 > 0

−4 > 0 no −3 > 0 no

3. log(𝑥2 + 7𝑥 + 12) − log(𝑥 + 4) = 0 4. log7(𝑥 − 2) + log7(𝑥 + 3) = log714

𝑥2+7𝑥+12

𝑥+4= 100 (𝑥 − 2)(𝑥 + 3) = 14

(𝑥+4)(𝑥+3)

(𝑥+4)= 1 𝑥2 + 𝑥 − 6 = 14

𝑥 + 3 = 1 𝑥2 + 𝑥 − 20 = 0

𝑥 = −2 √ (𝑥 − 4)(𝑥 + 5) = 0

Always check! 𝑥 = 4 & 𝑥 = −5(omit after checking)

5. ln((4𝑥)5) = 15

5 ln(4𝑥) = 15

ln4𝑥 = 3

𝑒3 = 4𝑥 𝑥 = 5.021384231 √ Always check!

34

6. log2((5𝑥 + 7)19) = 57

19log2(5𝑥 + 7) = 57

log2(5𝑥 + 7) = 3

5𝑥 + 7 = 23

5𝑥 + 7 = 8

5𝑥 = 1

𝑥 =1

5 √ Always check!

7. log(𝑥+2)(𝑥2 + 5𝑥 + 18) − log(𝑥+2)(2) = 2

𝑥2+5𝑥+18

2= (𝑥 + 2)2

𝑥2+5𝑥+18

2= (𝑥 + 2)(𝑥 + 2)

𝑥2+5𝑥+18

2= 𝑥2 + 4𝑥 + 4

2𝑥2 + 8𝑥 + 8 = 𝑥2 + 5𝑥 + 18

𝑥2 + 3𝑥 − 10 = 0

(𝑥 − 2)(𝑥 + 5) = 0

𝑥 = 2 & − 5 Always Check!

Omit −5 so only 𝑥 = 2 will work!

35

Homework 10.6: Solving Logarithmic Equations

1. Solve for 𝑥: ln(𝑥 + 2) + ln(𝑥 − 2) = ln(9𝑥 − 24)

2. Solve for 𝑥: log(10𝑥 + 5) − 3 = log (𝑥 − 5)

3. Solve for 𝑥: ln(32𝑥2) − 3 ln(2) = 3

4. Solve for 𝑥: log2(𝑥) + log2(2𝑥) + log2(3𝑥) + log2(36) = 6

5. Jenn claims that because log(1) + log(2) + log(3) = log (6), then

log(2) + log(3) + log(4) = log (9). Is she correct? Explain how you know.

36

Lesson 10.7: Change of Base Rule

Learning Goals:

1) What is the change of base rule for logarithms?

2) How do we solve logarithmic equations using change of base rule?

Warm Up:

Evaluate the following:

a) log28 use “logbase” in calc b) log418

2𝑥 = 8 4𝑥 = 18

𝑥log2 = log8 𝑥log4 = log18

𝑥 =log8

log2= 3 𝑥 =

log18

log4= 2.1

1. Let’s look at the two examples from the warm-up. Use the change of base

rule to rewrite the logarithms with base 10 and then use the LOG key in your

calculator.

a) log28 =log8

log2= 3 b) log418 =

log18

log4= 2.1

2. Use the change of base property to rewrite each logarithmic function in terms

of the common logarithm function.

a) 𝑔1(𝑥) = log14

(𝑥) b) 𝑔2(𝑥) = log12

(𝑥) c) 𝑔3(𝑥) = log2(𝑥)

log𝑥

log1

4

log𝑥

log1

2

log𝑥

log2

Change of Base Formula for Logarithms

If 𝑥, 𝑎, and 𝑏 are all positive real numbers with 𝑎 ≠ 1 and 𝑏 ≠ 1, then

log𝑏(𝑥) =log𝑎(𝑥)

log𝑎(𝑏)

37

3. Using the change of base rule, rewrite the given logarithmic expressions in

base 7.

a) log211 b. log5𝑥2 c. log𝑏(𝑥 + 3)

log711

log72

log7𝑥2

log75

log7(𝑥+3)

log7𝑏

Push yourself!

Show that for any positive numbers 𝑎 and 𝑏 with 𝑎 ≠ 1 and 𝑏 ≠ 1,

log𝑎(𝑏) ∙ log𝑏(𝑎) = 1.

logb

loga∙log𝑎

log𝑏= 1

1 = 1 √

How can we use the change of base formula to solve a logarithmic equation?

Rewrite the logarithmic expresssion with the larger base in terms of the

smaller base.

Cross multiply and drop the logs on both sides.

1. Solve for 𝑥: log(𝑥) = log100(𝑥2 − 2𝑥 + 6)

log𝑥

1=log (𝑥2−2𝑥+6)

log100

log𝑥

1=log (𝑥2−2𝑥+6)

2

2log𝑥 = log (𝑥2 − 2𝑥 + 6)

log𝑥2 = log (𝑥2 − 2𝑥 + 6) Cancel like bases!

𝑥2 = 𝑥2 − 2𝑥 + 6

2𝑥 = 6

𝑥 = 3 √ Always check! log(𝑥) = log100(𝑥2 − 2𝑥 + 6)

𝑥 > 0 (𝑥2 − 2𝑥 + 6) > 0

3 > 0 9 > 0

38

2. Solve for 𝑥: log2(𝑥 + 1) = log4⏟(𝑥2 + 3𝑥 + 4)

Change to base 2!

log2(𝑥 + 1) =log2(𝑥

2+3𝑥+4)

log24

log2(𝑥 + 1) =log2(𝑥

2+3𝑥+4)

2

2log2(𝑥 + 1) = log2(𝑥2 + 3𝑥 + 4)

log2(𝑥 + 1)2 = log2(𝑥

2 + 3𝑥 + 4) Cancel like bases!

(𝑥 + 1)2 = (𝑥2 + 3𝑥 + 4)

(𝑥 + 1)(𝑥 + 1) = 𝑥2 + 3𝑥 + 4

𝑥2 + 2𝑥 + 1 = 𝑥2 + 3𝑥 + 4

−𝑥 = 3

𝑥 = −3 Reject so “No Solution”!

3. Solve for 𝑥: log9⏟(𝑥2 + 2𝑥 + 6) = log3(𝑥 + 2)

Change to base 3!

log3(𝑥2+2𝑥+6)

log39= log3(𝑥 + 2)

log3(𝑥2+2𝑥+6)

2= log3(𝑥 + 2)

2log3(𝑥 + 2) = log3(𝑥2 + 2𝑥 + 6)

log3(𝑥 + 2)2 = log3(𝑥

2 + 2𝑥 + 6) Cancel like bases!

(𝑥 + 2)2 = 𝑥2 + 2𝑥 + 6

(𝑥 + 2)(𝑥 + 2) = 𝑥2 + 2𝑥 + 6

𝑥2 + 4𝑥 + 4 = 𝑥2 + 2𝑥 + 6

2𝑥 = 2 Always check! log9(𝑥2 + 2𝑥 + 6) = log3(𝑥 + 2)

𝑥 = 1 √ 𝑥2 + 2𝑥 + 6 > 0 𝑥 + 2 > 0

9 > 0 3 > 0

39

4. Solve for 𝑥: log(𝑥 − 2) = log100⏟ (14 − 𝑥)

Change to base 10!

log(𝑥 − 2) =log (14−𝑥)

log100

log(𝑥 − 2) =log (14−𝑥)

2

2 log(𝑥 − 2) = log (14 − 𝑥)

log(𝑥 − 2)2 = log (14 − 𝑥) Cancel like bases!

(𝑥 − 2)2 = 14 − 𝑥

(𝑥 − 2)(𝑥 − 2) = 14 − 𝑥

𝑥2 − 4𝑥 + 4 = 14 − 𝑥

𝑥2 − 3𝑥 − 10 = 0

(𝑥 − 5)(𝑥 + 2) = 0

𝑥 = 5 & − 2 Always check! log(𝑥 − 2) = log100(14 − 𝑥)

𝑥 − 2 > 0 14 − 𝑥 > 0

Must omit the −2 −4 > 0 16 > 0

So only 𝑥 = 5 √ 𝑥 − 2 > 0 14 − 𝑥 > 0

3 > 0 9 > 0

40

Homework 10.7: Change of Base Rule

1. Use the change of base rule to solve for 𝑥 in the given logarithmic equation:

log(𝑥 + 1) = log100(𝑥2 + 4𝑥)

2. Solve the given logarithmic equation: log4(𝑥) = log16(3𝑥2 − 3𝑥 + 1)

3. Which expression could be used to determine the value of 𝑦 in the equation

log6𝑥 = 𝑦? (1) log𝑥

6 (2)

log6

log𝑥 (3)

6

log𝑥 (4)

log𝑥

log6

4. Solve for 𝑥: ln(𝑥 + 2) + ln(𝑥 − 2) = ln (−2𝑥 − 1)

41

5. Solve for 𝑥: log (√(𝑥 + 3)3) =3

2

6. Drew said that the equation log2[(𝑥 + 1)4] = 8 cannot be solved because he

expanded (𝑥 + 1)4 = 𝑥4 + 4𝑥3 + 6𝑥2 + 4𝑥 + 1 and realized that he cannot solve

the equation 𝑥4 + 4𝑥3 + 6𝑥2 + 4𝑥 + 1 = 28. Is he correct? Explain how you

know.

7. Write ln (√5𝑥3

𝑒2) as a sum or difference of constants and logarithms of simpler

terms.