Lesson 1 Solving Logarithmic Equations

Lesson 1

Solving Logarithmic Equations

At the end of this lesson, the learner should be able to

• accurately apply the laws of logarithms on logarithmic equations;

• correctly solve logarithmic equations.

• correctly solve logarithmic inequalities; and

• correctly apply the laws and properties of logarithms in solving logarithmic inequalities.

Objectives

● How will you apply the laws of logarithms on logarithmic

equations?

● How will you solve logarithmic equations?

Essential Questions

Before we start with our lesson, let’s revisit the laws of logarithms and have a short drill about it by working as a team on the following online activity.

https://www.mathsisfun.com/algebra/exponents-logarithms.html

(Click the link posted in the chat box to access the Jamboardactivity.)

Warm-up!

● What are logarithms?

● What is the relationship between exponents and logarithms?

● What does the Laws of Logarithm state?

Guide Questions

Addition Law of Logarithms log𝑏 𝑚𝑛 = log𝑏𝑚 + log𝑏 𝑛

1

Example:

log2 5 ∙ 9 = log2 5 + log2 9

2 Logarithm of a Productthe logarithm of a product is equal to the sum of the logarithms of its factors

Example:

The logarithmic expression log 6 is equal to log 2 + log 3.

Subtraction Law of Logarithms log𝑏

𝑚

𝑛= log𝑏𝑚 − log𝑏 𝑛

3

Example:

log25

9= log2 5 − log2 9

4 Logarithm of a Quotient the logarithm of a quotient is equal to the logarithm of the numerator minus the logarithm of the denominator

Example:

The logarithmic expression log2

3is equal to log 2 − log 3.

Example:

log 34 = 4 log 3

5 Exponent Law of Logarithmslog𝑏𝑚

𝑛 = 𝑛 log𝑏𝑚

Example:

The logarithmic expression log 8 is equal to 3 log 2.

6 Logarithm of a Powerthe logarithm of a power 𝑚𝑛 to the base 𝑏 is equal to the product of the exponent 𝑛 and the logarithm of 𝑚 to the base 𝑏

Logarithmic Equationan equation that involves logarithmic expressions; the variable is written as part of the argument of the logarithm

7

Examples:

a. log3 𝑥 + 1 = 2b. log3 𝑥2 − 8 − log3 4 = 3c. ln 2𝑥 + 3 = ln 4𝑥

Rule Property

One-to-one correspondence If log𝑎 𝑥 = log𝑎 𝑦, then 𝑥 = 𝑦.

Identity property If log𝑎 𝑎 = 𝑦, then 𝑦 = 1.

Zero logarithm If log𝑎 1 = 𝑦, then 𝑦 = 0.

Some Properties of Logarithms8

Rule Property

Natural logarithm If ln 𝑥 = 𝑦, then log𝑒 𝑥 = 𝑦.

Equivalent exponential form If log𝑎 𝑥 = 𝑦, then 𝑎𝑦 = 𝑥.

Change of base If log𝑎 𝑥 = 𝑦 , then log𝑏 𝑥

log𝑏 𝑎= 𝑦.

Some Properties of Logarithms8

Example 1: Solve the logarithmic equation log2 𝑥 = 3.

Example 1: Solve the logarithmic equation log2 𝑥 = 3.

Solution:We use the definition of logarithms. Recall that if log𝑎 𝑥 = 𝑦,then 𝑎𝑦 = 𝑥.

log2 𝑥 = 3

𝑥 = 23

𝑥 = 8

Hence, the solution is 𝒙 = 𝟖.

Example 2: Find the solution(s) of the logarithmic equation

ln 3𝑥 − 2 + ln 4 = ln(𝑥 + 3).


ln 3𝑥 − 2 + ln 4 = ln(𝑥 + 3).

Solution:1. Apply the Addition Law of Logarithms on the left-hand

side of the equation.

ln 3𝑥 − 2 4 = ln(𝑥 + 3)


ln 3𝑥 − 2 + ln 4 = ln(𝑥 + 3).

Solution:2. Apply the one-to-one correspondence property of

logarithms.

3𝑥 − 2 4 = 𝑥 + 3


ln 3𝑥 − 2 + ln 4 = ln(𝑥 + 3).

Solution:3. Solve the resulting linear equation.

3𝑥 − 2 4 = 𝑥 + 312𝑥 − 8 = 𝑥 + 3

11𝑥 = 11𝑥 = 1


ln 3𝑥 − 2 + ln 4 = ln(𝑥 + 3).

Solution:Therefore, the solution is 𝒙 = 𝟏.

Logarithmic Inequalityan inequality that contains logarithmic expressions

9

Examples:

a. log4 𝑥 + 8 ≥ 11b. log(𝑥 − 2) < log 𝑥2 − 4c. 5 + ln 2𝑥 > 𝑒

One-to-One Correspondence Rule The following rules help in solving logarithmic inequalities.

a. log𝑏 𝑥 > log𝑏 𝑦 if and only if 𝑥 > 𝑦 and 𝑏 > 1.

b. log𝑏 𝑥 < log𝑏 𝑦 if and only if 𝑥 > 𝑦 and 0 < 𝑏 < 1.

10

Examples:

log2 32 > log2 16 since 32 > 16 and 𝑏 = 2 > 1.

Example 1: Solve the inequality log2(2𝑥 + 3) > log2 3𝑥.


Solution:We can apply the one-to-one correspondence rule in the inequality. Since log2(2𝑥 + 3) > log2 3𝑥, and the bases of the logarithms on each side of the inequality are equal and greater than 1, it follows that 𝟐𝒙 + 𝟑 > 𝟑𝒙.


Solution:Solve the resulting linear inequality.

2𝑥 + 3 > 3𝑥3 > 3𝑥 − 2𝑥3 > 𝑥


Solution:Therefore, 𝑥 < 3. However, since 2𝑥 + 3 and 3𝑥 are arguments in the original inequality, it means that 2𝑥 + 3 > 0and 3𝑥 > 0.

Solving for 𝑥, we get 𝑥 > −3

2and 𝑥 > 0.

Therefore, the solution set is 𝟎 < 𝒙 < 𝟑.

Example 2: Solve the inequality log3 𝑥 − 2 < 5.


Solution:1. Rewrite the right-hand side of the inequality as 𝐥𝐨𝐠𝟑 𝟑

𝟓 so that it becomes a logarithm with the same base as the logarithm on the left-hand side.

Note that log3 35 = 5 log3 3 = 5 1 = 5.

log3 𝑥 − 2 < log3 35


Solution:2. Apply the one-to-one correspondence rule on the

inequality.

Since log3(𝑥 − 2) < log3 35 and the bases of the logarithms on

each side of the inequality are equal and greater than 1, it follows that 𝒙 − 𝟐 < 𝟑𝟓.


Solution:Solve the resulting linear inequality.

𝑥 − 2 < 35

𝑥 − 2 < 243𝑥 < 245


Solution:Therefore, 𝑥 < 245. However, since 𝑥 − 2 is an argument in the original inequality, it means that 𝑥 − 2 > 0.

Solving for 𝑥, we get 𝑥 > 2.

Therefore, the solution set is 𝟐 < 𝒙 < 𝟐𝟒𝟓.

Individual Practice:

1. Solve the logarithmic equation log3(2𝑥 + 1) = 2.

2. Solve the logarithmic equation log3 𝑥 + log5 𝑥 = 3.

Group Practice: To be done in groups of four.

If a certain bacteria double in number every day, how long will it take for 1 000 bacteria to reach 16 384 000 in number?

Group Practice: To be done in groups of three or four.

Interns Maggie and Michelle are appointed to two different positions in a company. Maggie’s allowance (in thousand pesos) is log6(𝑥 − 2) while Michelle’s allowance (in thousand pesos) is log5(𝑥 − 2). Find the possible values of 𝑥 if Michelle’s allowance is higher than that of Maggie’s.

● What does the Laws of Logarithms state?

● How do you solve logarithmic equations?

● In which part of your daily life as a student can you see logarithmic equations?

● In what other fields can you apply logarithmic equations?

Synthesis

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Lesson 1 Solving Logarithmic Equations