Question

Find all real numbers $x$ which satisfies the equation $2{\log }_2{\log }_{2}x+{\log }_{\frac{1}{2}}{\log }_2\left(2\sqrt{2}x\right)=1$

• A. $8$
• B. $9$
• C. $10$
• D. $11$

Answer 1

The correct option is A $8$

${\log }_{2}x>0$

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

$\Rightarrow 2{\log }_2{\log }_{2}x-{\log }_2{\log }_2\left(2\sqrt{2}x\right)=1[\because {\log }_{\frac{1}{a}}b=-{\log }_{a}b]$

$\Rightarrow {\log }_2\left[\frac{{\left({\log }_{2}x\right)}^2}{{\log }_2\left(2\sqrt{2}x\right)}\right]=1[\because m\log a=\log a^m\text{\&}\log a-\log b=\log \frac{a}{b}]$

$\Rightarrow \frac{{\left({\log }_{2}x\right)}^2}{{\log }_2\left(2\sqrt{2}\right)+{\log }_{2}x}=2^1=2[\because \log (ab)=\log a+\log b]$

Substitute ${\log }_{2}x=t$ to obtain $t^2-2t-2{\log }_2\left(2\sqrt{2}\right)=0$

$\Rightarrow t^2-2t-3=0[\because 2{\log }_{2}2\sqrt{2}={\log }_2(2\sqrt{2}{)}^2={\log }_{2}8={\log }_2{2}^3=3]$

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

$\Rightarrow x=2^{-1}$or $2^3$

That is $x=\frac{1}{2}$ or $8$

Hence, the answer is $8$.