Question

Solve the following equation:
${\log }_{\sqrt{5}}x\cdot \sqrt{{\log }_{x}5\sqrt{5}+{\log }_{\sqrt{5}}5\sqrt{5}}=-\sqrt{6}$

Answer 1

$\log x\sqrt{5}.\sqrt{{\log }_x\sqrt[5]{55}+{\log }_{\sqrt{5}}\sqrt[5]{55}}=-\sqrt{6}\Rightarrow 2{\log }_{5}x\sqrt{{\frac{3}{2}\log }_{x}5+3}=-\sqrt{6}\Rightarrow 4\left({{\log }_{5}x)}^2\right[\frac{3}{2}{\log }_{x}5+3]=6\Rightarrow 6{\log }_{5}x+12{{\left(\log \right)}_{5}x}^2=6\Rightarrow 2({{\log }_{5}x)}^2+{\log }_{5}x-1=0$

Therefore, ${\log }_{5}x=\frac{1}{2},-1\Rightarrow x=\sqrt{5},\frac{1}{5}$

But $x=\sqrt{5}$ does not satisfy the given creation.

Therefore, $x=\frac{1}{5}$ is only possible solution.