Question

Find sum of all integer values of x which satisfy the equation $2{\log }_2{\log }_{2}x+{\log }_{1/2}{\log }_2\left(2\sqrt{2}x\right)=1$

Answer 1

$2{\log }_2{\log }_{2}x+{\log }_{1/2}{\log }_2\left(2\sqrt{2}x\right)=1$
$2{\log }_2{\log }_{2}x-{\log }_2{\log }_2\left(2\sqrt{2}x\right)=1$
${\log }_2({\log }_{2}x{)}^2-{\log }_2{\log }_2\left(2\sqrt{2}x\right)=1$
$\Rightarrow {\log }_2\left(\frac{({\log }_{2}x{)}^2}{{\log }_2\left(2\sqrt{2}x\right)}\right)=1$
$\Rightarrow 2=\frac{({\log }_{2}x{)}^2}{{\log }_2\left(2\sqrt{2}x\right)}$
$\Rightarrow ({\log }_{2}x{)}^2=2{\log }_2\left(2\sqrt{2}\right)+2{\log }_{2}x$
Put ${\log }_{2}x=t$
$\Rightarrow t^2-2t-{\log }_{2}8=0$
$\Rightarrow t^2-2t-3=0$
$\Rightarrow (t-3)(t+1)=0$
$\Rightarrow t=3,-1$
$\Rightarrow {\log }_{2}x=3,{\log }_{2}x=-1$
$\Rightarrow x=2^3,x=2^{-1}$
From the given eqn it follows that $x>0$
So, both values are possible