How do is solve this logarithm equation?
$$11 \cdot \log_3x+7 \cdot \log_7x = 13+3 \cdot \log_4x$$
I know that I have to use the change of base formula, but I still can't figure out the equation.
Can someone help me?
[Math] How to solve this logarithm equation with different bases
logarithms
Best Answer
The change of base formula is $$ \log_ab=\frac{\log b}{\log a} $$ where the base in the right hand side is whatever you prefer. I assume $e$. The equation becomes $$ \frac{11}{\log3}\log x+\frac{7}{\log7}\log x=13+\frac{3}{\log4}\log x $$ which is a first degree equation in $\log x$.