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

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

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

$⟹2{\log }_2\left({\log }_{2}x\right)-{\log }_2(\frac{3}{2}+{\log }_{2}x)=1$

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

$⟹2{\log }_{2}m-{\log }_{2}m=1+{\log }_2\frac{3}{2}={\log }_{2}2+{\log }_2\frac{3}{2}$

$⟹{\log }_{2}m={\log }_2\left(2.\frac{3}{2}\right)={\log }_{2}3$

$⟹m=3$

$⟹{\log }_{2}x=3$

$⟹x=2^3$

$⟹x=8$