Question

Solve for $x$: $x^{{\log }_{4}x}=2^{3({\log }_{4}x+3)}$

Answer 1

$x^{{\log }_{4}x}=2^{3({\log }_{4}x+3)}$
Taking logarithm (with base 2) on both sides, we get

${\log }_2{x}^{{\log }_{4}x}={\log }_2{2}^{3({\log }_{4}x+3)}$
${\log }_{4}x{\log }_{2}x=3({\log }_{4}x+3)$ ($\because \log x^m=m\log x;{\log }_{a}a=1$)
$\frac{1}{2}{\log }_{2}x{\log }_{2}x=\frac{3}{2}{\log }_{2}x+9$ ($\because {\log }_{a^b}x=\frac{1}{b}{\log }_{a}x$)
$\Rightarrow {\left({\log }_{2}x\right)}^2-3{\log }_{2}x-18=0$
Put ${\log }_{2}x=t$
$\Rightarrow t^2-3t-18=0$
$\Rightarrow (t-6)(t+3)=0$
$\Rightarrow t=6,-3$
$\Rightarrow {\log }_{2}x=6,{\log }_{2}x=-3$

Converting into exponential form
$\Rightarrow x=2^6,x=\frac{1}{8}$