# Solving Logarithmic Equations

Do you want to know how to solve Logarithmic Equations? you can do it in two easy steps.

## Step by step guide to solve Logarithmic Equations

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

### Example 1:

Find the value of the variables in each equation. $$\log_{4}{(20-x^2)}=2$$

Use log rule: $$\log_{b}{x}=\log_{b}{y}$$, then: $$x=y$$

$$2=\log_{4}{4^2},\log_{4}{(20-x^2)}=\log_{4}{4^2}=\log_{4}{16}$$

then: $$20-x^2=16→20-16=x^2→x^2=4→x=2$$

### Example 2:

Find the value of the variables in each equation. $$log⁡(2x+2)=log⁡(4x-6)$$

When the logs have the same base: $$?(?)=?(?)$$,?ℎ??: $$??(?(?))=??(?(?)) ???2?+2=???4?−6→2?+2=4?−6→2?+2−4?+6=0$$

$$2x+2-4x+6=0→-2x+8=0→-2x=-8→x=\frac{-8}{-2}=4$$

### Example 3:

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

Use log rule: $$\log_{b}{x}=\log_{b}{y}$$, then: $$x=y$$

$$2=\log_{2}{2^2},\log_{2}{(25-x^2)}=\log_{2}{2^2}=\log_{2}{4}$$

Then: $$25-x^2=4→25-16=x^2→x^2=9→x=3$$

### Example 4:

Find the value of the variables in each equation. $$log⁡(8x+3)=log⁡(2x-6)$$

When the logs have the same base: $$?(?)=?(?)$$,?ℎ??: $$??(?(?))=??(?(?))$$

$$log⁡(8x+3)=log⁡(2x-6)→8x+3=2x-6→8x+3-2x+6=0$$

$$6x+9=0→6x=-9→x=\frac{-9}{6}=-\frac{3}{2}$$

## Exercises

### Find the value of the variables in each equation.

1. $$log⁡(x+5)=2$$
2. $$log x-log 4=3$$
3. $$log x+log 2=4$$
4. $$log 10+log x=1$$
5. $$log x+log 8=log 48$$
6. $$-3\log_{3}{(x-2)}=-12$$
7. $$log 6x=log (x+5)$$
8. $$log (4k-5)=log (2k-1)$$