Question

${\log }_{\sqrt{5}}x.\sqrt{{\log }_{x}5\sqrt{5}+{\log }_{\sqrt{5}}}5\sqrt{5}=-\sqrt{6}$

Answer 1

Let ${\log }_{5}x=t$

$\therefore {\log }_{5^{1/2}}x.\sqrt{{\log }_x{5}^{3/2}+{\log }_{5^{1/2}}{5}^{3/2}}=-\sqrt{6}$

$\Rightarrow 2{\log }_{5}x.\sqrt{\frac{3}{2}{\left({\log }_{5}x\right)}^{-1}+3}=-\sqrt{6}$

$=2t.\sqrt{3}.\sqrt{\frac{1}{2t}+1}=-\sqrt{2}.\sqrt{3}$

$\Rightarrow t\sqrt{\frac{1}{2t}+1}=\frac{-1}{\sqrt{2}}$

$\Rightarrow \sqrt{\frac{t}{2}+t^2}=\frac{-1}{\sqrt{2}}\Rightarrow t^2+\frac{t}{2}=\frac{1}{2}$

$\Rightarrow 2t^2+t-1=0\Rightarrow (2t-1)(t+1)=0$

$\Rightarrow t=\frac{1}{2}$ or $t=-1$

$\Rightarrow {\log }_{5}x=\frac{1}{2}$ or ${\log }_{5}x=-1$

$\Rightarrow x=\sqrt{5}$ or, $x=\frac{1}{5}$