# How to Solve Logarithmic Equations

In this blog post, using the definition and rules of logarithms (exponentials and change-of-base formula), you will be taught how to solve logarithmic equations.

## A step-by-step guide to solving Logarithmic Equations

• Convert the logarithmic equation to an exponential equation when it’s possible. (If no base is indicated, the base of the logarithm is 10)
• Condense logarithms if you have more than one log on one side of the equation.
• Plugin the answers back into the original equation and check to see if the solution works.

## Examples

### Solving Logarithmic Equations – Example 1:

Find the value of $$x$$ in this equation. $$log_{ 2}{(36-x^2)}=4$$

Solution:

Use log rule: $$log_{ b}{x}=log_{b}{ y}$$, then: $$x=y$$
$$4=log_{2} {(2^4)} , log_{ 2}{(36-x^2 )}=log_{ 2}{(2^4 )}=log_{2}{ 16}$$
Then: $$36-x^2=16→36-16=x^2→x^2=20→x=\sqrt{20}=2\sqrt{5}$$

### Solving Logarithmic Equations – Example 2:

Find the value of $$x$$ in this equation. $$log⁡(5x+2)=log⁡(3x-1)$$

Solution:

When the logs have the same base: $$f(x)=g(x)$$,then: $$ln(f(x))=ln(g(x))$$, $$log⁡(5x+2)=log⁡(3x-1)→5x+2=3x-1→5x+2-3x+1=0→2x+3=0→2x=-3→x=-\frac{3}{2}$$
Verify Solution: $$log(5x+2)=log⁡(5(-\frac{3}{2})+2)=log⁡(-5.5)$$
Logarithms of negative numbers are not defined. Therefore, there is no solution for this equation.

### Solving Logarithmic Equations – Example 3:

Find the value of the variables in this equation. $$log_{2}{(25-x^2)}=4$$

Solution:

Use the logarithmic definition: $$log_{a⁡}{(b}=c→a^c=b$$
$$log_{2}{(25-x^2 )}=4→2^4=(25-x^2 )→16=(25-x^2 )$$
Simplify: $$16=(25-x^2 )→-x^2+25-16=0$$
Then: $$x^2=9→x=3$$ or $$-3$$ Both 3 and $$-3$$ work in the original equation.

### Solving Logarithmic Equations – Example 4:

Find the value of $$x$$ in this equation. $$log⁡(3x+10)=log⁡(6x-8)$$

Solution:

When the logs have the same base: $$f(x)=g(x)$$,then: $$ln(f(x))=ln(g(x))$$, $$log⁡(3x+10)=log⁡(6x-8)→3x+10=6x-8→3x+10-10=6x-8-10→3x=6x-18→3x-6x=6x-18-6x→-3x=-18→x=\frac{-18}{-3}=6$$
Verify Solution: $$log(3x+10)=log⁡(3(6)+10)=log⁡(28)$$
Logarithms of negative numbers are not defined. Therefore, there is no solution for this equation.

## Exercises for Solving Logarithmic Equations

### Solve Logarithmic Equations.

1. $$\color{blue}{log_{5}{3x}=0,x=}$$
2. $$\color{blue}{log_{2}{6x}=2,x=}$$
3. $$\color{blue}{(logx)+3=1,x=}$$
4. $$\color{blue}{log3x=log(x+1),x=}$$
5. $$\color{blue}{log2-logx=0,x=}$$
6. $$\color{blue}{log(2x-1)=log(4x-2),x=}$$
1. $$\color{blue}{\frac{1}{3}}$$
2. $$\color{blue}{\frac{2}{3}}$$
3. $$\color{blue}{\frac{1}{100}}$$
4. $$\color{blue}{\frac{1}{2}}$$
5. $$\color{blue}{2}$$
6. $$\color{blue}{No \ solution \ for \ x∈R}$$

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