Exponential functions and logarithms: Logarithmic functions
More logarithmic equations
How do you solve the following equation?
\[2\cdot\log_{3}\left(x\right)=1+\log_{3}\left(3 x-6\right)\]
Drag the steps in the correct order.
- Step 1
- Step 2
- Step 3
- Step 4
- Step 5
- Step 6
- Step 7
- Write the equation #\green{b}=\purple{c}#
- Take logarithms together so you get an equation of the form #\log_\blue{a}\left(\green{b}\right)=\log_\blue{a}\left(\purple{c}\right) #
- Rewrite the equation #\green{b}=\purple{c}# as an equation you can solve using the quadratic formula
- Solve the equation using the quadratic formula
- Check if the solutions are valid
- Write all terms as a logarithm
- Apply #\purple{n}\cdot \log_{\blue{a}}\left(\green{b}\right)=\log_{\blue{a}}\left(\green{b}^\purple{n}\right) #
- #2\cdot\log_{3}\left( x\right)=\log_3\left( 3\right)+\log_{3}\left( 3 x-6\right) #
- #\log_{3}\left( x^2\right)=\log_{3}\left( 9 x-18\right) #
- #x^2-9x+18=0#
- #\log_{3}\left( x^2\right)=\log_3\left( 3\right)+\log_{3}\left( 3 x-6\right) #
- #x^2=9 x-18#
- #x=3\vee x=6#
- Both values of #x# yield positive expressions within the logarithms, so they are both solutions to the equation
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