Question

The solution set of
$2{\log }_2{\log }_{2}x+{\log }_{1/2}{\log }_2\left(2\sqrt{2x}\right)=1\text{is}$

Answer 1

$\text{Given}:2{\log }_2{\log }_{2}x+{\log }_{1/2}{\log }_2\left(2\sqrt{2x}\right)=1$
$\Rightarrow {\log }_{2}2{\log }_{2}x+{\log }_{1/2}{\log }_2\left(2\sqrt{2x}\right)=1\Rightarrow {\log }_2{\log }_2{x}^2+{\log }_{1/2}{\log }_2(8x{)}^{\frac{1}{2}}=1\Rightarrow {\log }_2{\log }_2{x}^2+{\log }_{1/2}(\frac{1}{2}\cdot {\log }_2(8x))=1\Rightarrow {\log }_2{\log }_2{x}^2+{\log }_{1/2}\frac{1}{2}+{\log }_{1/2}{\log }_{2}8x=1\Rightarrow {\log }_2{\log }_2{x}^2+1+{\log }_{1/2}{\log }_{2}8x=1\Rightarrow {\log }_2{\log }_2{x}^2=-{\log }_{1/2}{\log }_{2}8x\Rightarrow {\log }_2{\log }_2{x}^2={\log }_2{\log }_{2}8x\Rightarrow x^2=8x\Rightarrow x^2-8x=0\Rightarrow x(x-8)=0x=8,0\text{(not possible,because}x>0)$
Hence, The correct answer is option (a)