Question

Solve:-
$2{\log }_2\left({\log }_{2}x\right)+{\log }_{\frac{1}{2}}\left({\log }_{2}2\sqrt{2}x\right)=1$

Answer 1

$2{\log }_2\left({\log }_{2}x\right)+{\log }_{\frac{1}{2}}\left({\log }_{2}2\sqrt{2}x\right)=1$

$\Rightarrow {\log }_2{\left({\log }_{2}x\right)}^2-{\log }_2\left({\log }_{2}2\sqrt{2}x\right)=1$

$\Rightarrow {\log }_2\left(\frac{{\left({\log }_{2}x\right)}^2}{\left({\log }_{2}2\sqrt{2}x\right)}\right)=1$

$\Rightarrow \frac{{\left({\log }_{2}x\right)}^2}{\left({\log }_{2}2\sqrt{2}x\right)}=2$

$\Rightarrow {\left({\log }_{2}x\right)}^2=2\left({\log }_{2}2\sqrt{2}x\right)$

$\Rightarrow {\left({\log }_{2}x\right)}^2=2({\log }_2{2}^{1+\frac{1}{2}}+{\log }_{2}x)$

$\Rightarrow {\left({\log }_{2}x\right)}^2=2({\log }_2{2}^{\frac{3}{2}}+{\log }_{2}x)$

$\Rightarrow {\left({\log }_{2}x\right)}^2=2(\frac{3}{2}{\log }_{2}2+{\log }_{2}x)$

$\Rightarrow {\left({\log }_{2}x\right)}^2=3+2{\log }_{2}x$

Let ${\log }_{2}x=\alpha$

$\Rightarrow {\alpha }^2-2\alpha -3=0$

$\Rightarrow {\alpha }^2-3\alpha +\alpha -3=0$

$\Rightarrow \alpha (\alpha -3)+1(\alpha -3)=0$

$\Rightarrow (\alpha -3)(\alpha +1)=0$

$\Rightarrow \alpha =-1,3$

$\Rightarrow {\log }_{2}x=-1,3$

$\therefore x=2^{-1},2^3$

$\therefore x=\frac{1}{2},8$