Question

Solve the following equation.
${\log }_{\sqrt{5}}(4x^x-6)-{\log }_{\sqrt{5}}(2^x-2)=2$

Answer 1

${log}_{\sqrt{5}}(4^x-6)-{log}_{\sqrt{5}}(2^x-2)=2\Rightarrow {log}_{\sqrt{5}}\left(\frac{4^x-6)}{2^x-2}\right)=2\Rightarrow \frac{4^x-6}{2^x-2}=5\Rightarrow 4^x-6={5.2}^x-10\Rightarrow 4^x-{5.2}^x+4=0$

Therefore, $2^x=1,4\Rightarrow x=0,2$

At $x=0$, given equation is not valid.

Therefore, $4^x-6<0.$

Therefore, $x=2$ is the solution.