How to Solve Logarithmic Equations? (+FREE Worksheet!)
In this blog post, you will learn how to solve Logarithmic Equations using the properties of logarithms in a few easy steps.

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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.
- Plug the answers back into the original equation and check if the solution works.
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Logarithmic Equations – 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: \(20-x^2=16→20-16=x^2→x^2=4→x=2\) or \(x=-2\)
Logarithmic Equations – 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: \(f(x)=g(x)\),then: \(ln(f(x))=ln(g(x))\),
\(log(2x+2)=log(4x-6)→2x+2=4x-6→2x+2-4x+6=0\)
\(2x+2-4x+6=0→-2x+8=0→-2x=-8→x=\frac{-8}{-2}=4\)
Logarithmic Equations – 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: \(25-x^2=4→25-4=x^2→x^2=21 →x=\sqrt{21} \) or \(-\sqrt{21}\)
Logarithmic Equations – 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: \(f(x)=g(x)\),then: \(ln(f(x))=ln(g(x))\),
\(log(8x+3)=log(2x-6)→8x+3=2x-6→8x+3-2x+6=0\)
\(6x+9=0→6x=-9→x=\frac{-9}{6}=-\frac{3}{2}\)
Logarithms of negative numbers are not defined. Therefore, there is no solution for this equation.
Exercises for Logarithmic Equations
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Find the value of the variables in each equation.
- \(\color{blue}{log(x+5)=2}\)
- \(\color{blue}{log x-log 4=3}\)
- \(\color{blue}{log x+log 2=4}\)
- \(\color{blue}{log 10+log x=1}\)
- \(\color{blue}{log x+log 8=log 48}\)
- \(\color{blue}{-3\log_{3}{(x-2)}=-12}\)
- \(\color{blue}{log 6x=log (x+5)}\)
- \(\color{blue}{log (4k-5)=log (2k-1)}\)

Answers
- \(\color{blue}{95}\)
- \(\color{blue}{4000}\)
- \(\color{blue}{5000}\)
- \(\color{blue}{1}\)
- \(\color{blue}{6}\)
- \(\color{blue}{83}\)
- \(\color{blue}{1}\)
- \(\color{blue}{2}\)
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