Question

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

Answer 1

$2{\log }_2{\log }_{2}x+{\log }_{\frac{1}{2}}\left({\log }_2\left(2\sqrt{2}x\right)\right)=1$
Domain: $x>1$
$\Rightarrow 2{\log }_2{\log }_{2}x-{\log }_2{\log }_2\left(2\sqrt{2}x\right)=1[\because {\log }_{1/a}b=-{\log }_{a}b]$
$\Rightarrow {\log }_2\left(\frac{{\left({\log }_{2}x\right)}^2}{{\log }_{2}2\sqrt{2}x}\right)=1[\because \log a-\log b=\log \frac{a}{b}]$
$\Rightarrow {\left({\log }_{2}x\right)}^2=2{\log }_{2}2\sqrt{2}x$
$\Rightarrow {\left({\log }_{2}x\right)}^2=2{\log }_{2}2\sqrt{2}x$
$\Rightarrow {\left({\log }_{2}x\right)}^2={\log }_2\left(8x^2\right)=3+2{\log }_{2}x[\because \log (ab)=\log a+\log b\text{\&}{\log }_{a}a=1]$
$\Rightarrow y^2-2y-3=0,$ where ${\log }_{2}x=y$
$\Rightarrow y=3;y=-1$
$\Rightarrow x=8;x=\frac{1}{2}$
Due to domain constraint, only $x=8$ satisfies the equation.