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

- 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.
- 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. \(\log_{4}{(20-x^2)}=2\)

**Solution:**

*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. \(log(2x+2)=log(4x-6)\)

**Solution:**

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. \(\log_{2}{(25-x^2)}=2\)

**Solution:**

*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. \(log(8x+3)=log(2x-6)\)

**Solution: **

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.

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

## Answers

- {-1/1,000}
- {4,000}
- {5,000}
- {1}
- {6}
- {83}
- {1}
- {2}