Question

Find $x$ satisfying the expression ${\log }_4{\log }_{2}x+{\log }_2{\log }_{4}x=1$

Answer 1

$\log x$ will have real solutions when $x>0$.
${\log }_4{\log }_{2}x+{\log }_2{\log }_{4}x=2$

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

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

$\Rightarrow \frac{1}{2}{\log }_2{\log }_{2}x+{\log }_2{\log }_{2}x-1=1[\because \log a^m=m\log a\text{\&}{\log }_{a}a=1]$

$\Rightarrow ({\log }_2{\log }_{2}x)=2$

$\Rightarrow {\log }_{2}x=2^2=4$

$\Rightarrow x=4^2=16$
Ans: 16