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# Solving Logarithmic Equations â€“ 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, letâ€™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.Â The first law is represented as;

âŸ¹ 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.

âŸ¹ log _{b }(x) â€“ log _{b }(y) = log (x/y)

**The power rule:**

âŸ¹ log _{b }(x) ^{n} = n log_{ b} (x)

*Change of base rule.*

âŸ¹ log _{b }xÂ = (log _{a }x)Â / (log _{a }b)

**Identity rule**

The logarithm of any positive number to the same base of that number is always 1.

b^{1}=b âŸ¹ log _{b }(b)=1.

Example:

- The logarithm of the number 1 to any non-zero base is always zero.

b^{0}=1 âŸ¹ 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:*

- Equations containing logarithms on one side of the equation.
- 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 â‡’ M = b ^{n}.

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

- Simplify the logarithmic equations by applying the appropriate laws of logarithms.
- RewriteÂ the logarithmic equation inÂ 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 â‡’ 2 ^{5 }= 5x + 7

â‡’ 32 = 5x + 7

â‡’ 5x = 32 â€“ 7

5x = 25

Divide both sides by 5 to get

x = 5

*Example 2*

Solve for x inÂ 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.

â‡’ 10^{2 }= 5x â€“ 11

â‡’ 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

log_{10}Â (2x + 1) = 3nâ‡’ 2x + 1 = 10^{3}

â‡’ 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;

â‡’ log_{10}Â (2 x 499.5 + 1)Â = log_{10}Â (1000) = 3 since 10^{3}Â = 1000

*Example 4*

Evaluate ln (4x -1) = 3

__Solution__

RewriteÂ theÂ equation inÂ exponential form as;

ln (4x -1) = 3 â‡’ 4x â€“ 3 =e^{3}

But as you know, e = 2.718281828

4x â€“ 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 â‡’ log _{2 }[(x + 1)/ (x â€“ 4)] = 3

Now, rewrite the equation in exponential form

â‡’2 ^{3 }= [(x + 1)/ (x â€“ 4)]

â‡’ 8 = [(x + 1)/ (x â€“ 4)]

Cross multiply the equation

â‡’ [(x + 1) = 8(x â€“ 4)]

â‡’ x + 1 = 8x -32

7x = 33 â€¦… (Collecting the like terms)

x = 33/7

*Example 6*

SolveÂ 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 â‡’ log _{4 }[(x) (x â€“ 12)] = 3

â‡’ log _{4 }(x^{2 }– 12x) = 3

Convert the equation in exponential form.

â‡’ 4^{3 =} x^{2 }– 12x

â‡’ 64 = x^{2 }– 12x

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

x^{2} -12x â€“ 64 â‡’ (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 â€“ 4) + log _{6 (}4) = log_{ 6 }(40)

__Solution__

First, simplify the logarithms.

log _{6 }(2x â€“ 4) + log _{6 }(4) = log_{ 6 }(40) â‡’ log _{6 }[4(2x â€“ 4)] = log _{6 }(40)

Now drop the logarithms

â‡’ [4(2x â€“ 4)] = (40)

â‡’ 8x â€“ 16 = 40

â‡’ 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.

â‡’ [(x – 2) (x + 3)] = 14

Distribute the FOIL to get;

â‡’ x ^{2 }– x â€“ 6 = 14

â‡’ x ^{2 }â€“ x â€“ 20 = 0

â‡’ (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}Â x + log _{3}Â (x + 3)Â =Â log _{3}Â (2x + 6)

__Solution__

Given the equation; log _{3}Â (x^{2}Â + 3x)Â = log _{3}Â (2x + 6), drop the logarithms to get;

â‡’ x^{2}Â + 3x = 2x + 6

â‡’ x^{2}Â + 3x – 2x – 6 = 0

x^{2}Â + x – 6 = 0â€¦â€¦â€¦â€¦â€¦â€¦ (Quadratic equation)

Factor the quadratic equation to get;

(x – 2) (x + 3) = 0

x = 2 and xÂ = -3

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

*Example 10*

Solve log _{5}Â (30x – 10) – 2 = log _{5}Â (x + 6)

__Solution__

log _{5}Â (30x – 10) – 2 = log _{5}Â (x + 6)

This equation can be rewritten as;

â‡’ log _{5}Â (30x – 10) –Â log _{5}Â (x + 6) = 2

Simplify the logarithms

log _{5}Â [(30x â€“ 10)/ (x + 6)] = 2

Rewrite logarithm in exponential form.

â‡’ 5^{2 }= [(30x â€“ 10)/ (x + 6)]

â‡’ 25 = [(30x â€“ 10)/ (x + 6)]

On cross multiplying, we get;

â‡’ 30x – 10 = 25 (x + 6)

â‡’ 30x – 10 = 25x + 150

â‡’ 30x – 25x = 150 + 10

â‡’ 5x = 160

x = 32