Question

Solve the following equation:
${\log }_4(x+12){\log }_{x}2=1$

Answer 1

${\log }_4(x+12){\log }_{x}2=1\Rightarrow \frac{\log (x+12)}{2\log 2}.\frac{\log 2}{\log x}=1\Rightarrow \frac{\log (x+12)}{2\log x}=1\Rightarrow x+12=x^2\Rightarrow x^2-x-12=0\Rightarrow x=4,-3.$

But $x$ cannot be $-3$.

Therefore, $x=4$ is the solution.