Question

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

Answer 1

We have,

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

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

We have,

${\log }_{a^n}x=\frac{1}{n}{\log }_{a}x$

Therefore

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

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

We know that,

$\log m-\log n=\log \left(\frac{m}{n}\right)$

Therefore,

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

We know that

${\log }_{a}x=n$

$x=a^n$

Therefore,

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

$({\log }_{2}x{)}^2=2{\log }_{2}2\sqrt{2}$

$({\log }_{2}x{)}^2={\log }_2{\left(2\sqrt{2}\right)}^2$

$({\log }_{2}x{)}^2={\log }_{2}8$

$({\log }_{2}x{)}^2=3{\log }_{2}2$

$({\log }_{2}x{)}^2=3$

${\log }_{2}x=\sqrt{3}$

$x=2^{\sqrt{3}}$

Hence, the value is $x$ is $2^{\sqrt{3}}$.