 # Solving Logarithmic Equations – Explanation & Examples

As you well know that, a logarithm is a mathematical operation that is the inverse of exponentiation. The logarithm of a number is abbreviated as “log.”

Before we can get into solving logarithmic equations, let’s first familiarize ourselves with the following rules of logarithms:

• The product rule:

The product rule states that the sum of two logarithms is equal to the product of the logarithms.  The first law is represented as;

⟹ log b (x) + log b (y) = log b (xy)

• The quotient rule:

The difference of two logarithms x and y is equal to the ratio of the logarithms.

⟹ log b (x) – log b (y) = log (x/y)

• The power rule:

⟹ log b (x) n = n log b (x)

• Change of base rule.

⟹ log b x = (log a x) / (log a b)

• Identity rule

The logarithm of any positive number to the same base of that number is always 1.
b1=b ⟹ log b (b)=1.

Example:

• The logarithm of the number 1 to any non-zero base is always zero.
b0=1 ⟹ log b 1 = 0.

## How to Solve Logarithmic Equations?

An equation containing variables in the exponents is knowns as an exponential equation. In contrast, an equation that involves the logarithm of an expression containing a variable is referred to as a logarithmic equation.

The purpose of solving a logarithmic equation is to find the value of the unknown variable.

In this article, we will learn how to solve the general two types of logarithmic equations, namely:

1. Equations containing logarithms on one side of the equation.
2. Equations with logarithms on opposite sides of the equal to sign.

### How to solve equations with logarithms on one side?

Equations with logarithms on one side take log b M = n ⇒ M = b n.

To solve this type of equations, here are the steps:

• Simplify the logarithmic equations by applying the appropriate laws of logarithms.
• Rewrite the logarithmic equation in exponential form.
• Now simplify the exponent and solve for the variable.
• Verify your answer by substituting it back in the logarithmic equation. You should note that the acceptable answer of a logarithmic equation only produces a positive argument.

Example 1

Solve log 2 (5x + 7) = 5

Solution

Rewrite the equation to exponential form

logs 2 (5x + 7) = 5 ⇒ 2 5 = 5x + 7

⇒ 32 = 5x + 7

⇒ 5x = 32 – 7

5x = 25

Divide both sides by 5 to get

x = 5

Example 2

Solve for x in log (5x -11) = 2

Solution

Since the base of this equation is not given, we therefore assume the base of 10.

Now change the write the logarithm in exponential form.

⇒ 102 = 5x – 11

⇒ 100 = 5x -11

111= 5x

111/5 = x

Hence, x = 111/5 is the answer.

Example 3

Solve log 10 (2x + 1) = 3

Solution

Rewrite the equation in exponential form

log10 (2x + 1) = 3n⇒ 2x + 1 = 103

⇒ 2x + 1 = 1000

2x = 999

On dividing both sides by 2, we get;

x = 499.5

Verify your answer by substituting it in the original logarithmic equation;

⇒ log10 (2 x 499.5 + 1) = log10 (1000) = 3 since 103 = 1000

Example 4

Evaluate ln (4x -1) = 3

Solution

Rewrite the equation in exponential form as;

ln (4x -1) = 3 ⇒ 4x – 3 =e3

But as you know, e = 2.718281828

4x – 3 = (2.718281828)3 = 20.085537

x = 5.271384

Example 5

Solve the logarithmic equation log 2 (x +1) – log 2 (x – 4) = 3

Solution

First simplify the logarithms by applying the quotient rule as shown below.

log 2 (x +1) – log 2 (x – 4) = 3 ⇒ log 2 [(x + 1)/ (x – 4)] = 3

Now, rewrite the equation in exponential form

⇒2 3 = [(x + 1)/ (x – 4)]

⇒ 8 = [(x + 1)/ (x – 4)]

Cross multiply the equation

⇒ [(x + 1) = 8(x – 4)]

⇒ x + 1 = 8x -32

7x = 33 …… (Collecting the like terms)

x = 33/7

Example 6

Solve for x if log 4 (x) + log 4 (x -12) = 3

Solution

Simplify the logarithm by using the product rule as follows;

log 4 (x) + log 4 (x -12) = 3 ⇒ log 4 [(x) (x – 12)] = 3

⇒ log 4 (x2 – 12x) = 3

Convert the equation in exponential form.

⇒ 43 = x2 – 12x

⇒ 64 = x2 – 12x

Since this is a quadratic equation, we therefore solve by factoring.

x2 -12x – 64 ⇒ (x + 4) (x – 16) = 0

x = -4 or 16

When x = -4 is substituted in the original equation, we get a negative answer which is imaginary. Therefore, 16 is the only acceptable solution.

### How to solve equations with logarithms on both sides of the equation?

The equations with logarithms on both sides of the equal to sign take log M = log N, which is the same as M = N.

The procedure of solving equations with logarithms on both sides of the equal sign.

• If the logarithms have are a common base, simplify the problem and then rewrite it without logarithms.
• Simplify by collecting like terms and solve for the variable in the equation.
• Check your answer by plugging it back in the original equation. Remember that, an acceptable answer will produce a positive argument.

Example 7

Solve log 6 (2x – 4) + log 6 (4) = log 6 (40)

Solution

First, simplify the logarithms.

log 6 (2x – 4) + log 6 (4) = log 6 (40) ⇒ log 6 [4(2x – 4)] = log 6 (40)

Now drop the logarithms

⇒ [4(2x – 4)] = (40)

⇒ 8x – 16 = 40

⇒ 8x = 40 + 16

8x= 56

x = 7

Example 8

Solve the logarithmic equation: log 7 (x – 2) + log 7 (x + 3) = log 7 14

Solution

Simplify the equation by applying the product rule.

Log 7 [(x – 2) (x + 3)] = log 7 14

Drop the logarithms.

⇒ [(x – 2) (x + 3)] = 14

Distribute the FOIL to get;

⇒ x 2 – x – 6 = 14

⇒ x 2 – x – 20 = 0

⇒ (x + 4) (x – 5) = 0

x = -4 or x = 5

when x = -5 and x = 5 are substituted in the original equation, they give a negative and positive argument respectively. Therefor, x = 5 is the only acceptable solution.

Example 9

Solve log 3 x + log 3 (x + 3) = log 3 (2x + 6)

Solution

Given the equation; log 3 (x2 + 3x) = log 3 (2x + 6), drop the logarithms to get;
⇒ x2 + 3x = 2x + 6
⇒ x2 + 3x – 2x – 6 = 0
x2 + x – 6 = 0……………… (Quadratic equation)
Factor the quadratic equation to get;

(x – 2) (x + 3) = 0
x = 2 and x = -3

By verifying both values of x, we get x = 2 to be the correct answer.

Example 10

Solve log 5 (30x – 10) – 2 = log 5 (x + 6)

Solution

log 5 (30x – 10) – 2 = log 5 (x + 6)

This equation can be rewritten as;

⇒ log 5 (30x – 10) –  log 5 (x + 6) = 2

Simplify the logarithms

log 5 [(30x – 10)/ (x + 6)] = 2

Rewrite logarithm in exponential form.

⇒ 52 = [(30x – 10)/ (x + 6)]

⇒ 25 = [(30x – 10)/ (x + 6)]

On cross multiplying, we get;

⇒ 30x – 10 = 25 (x + 6)

⇒ 30x – 10 = 25x + 150

⇒ 30x – 25x = 150 + 10

⇒ 5x = 160

x = 32

### Practice Questions

1. Which of the following shows the solution/s to $\log _3\left(5x+1\right)=4$?

2. Which of the following shows the solution/s to $\log _{10}\left(25x+400\right)=3$?

3. Which of the following shows the solution/s to $\log _{10}\left(3x+1\right) = 2$?

4. Which of the following shows the solution/s to $\ln \left(2x-3\right) = 2$?

5. Which of the following shows the solution/s to $\log _3\left(x+2\right)-\log _3\left(x-2\right) = 2$?

6. Which of the following shows the solution/s to $\log _2\left(x\right)+\log _2\left(x-4\right) = 1$?

7. Which of the following shows the solution/s to $\log _4\left(x-1\right)+\log _4\left(8\right)=\log _4\left(36\right)$?

8. Which of the following shows the solution/s to $\log _8\left(x-2\right)+\log _8\left(x+2\right)=\log _8\left(16\right)$?

9. Which of the following shows the solution/s to $\log _8\left(x-2\right)+\log _8\left(x+2\right)=\log _8\left(16\right)$?

10. Which of the following shows the solution/s to $\log _4\left(20x-15\right)-1=\log _4\left(x+2\right)$?

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