Question

Find $x^{\log x+4}=32$ where base of logarithm is $2$

Answer 1

$x^{{\log }_{2}x+4}=2^5$

Applying ${\log }_2$ to both sides,

$\Rightarrow ({\log }_{2}x+4){\log }_{2}x=5{\log }_{2}2$

$\Rightarrow ({\log }_{2}x+4){\log }_{2}x=5$

$\Rightarrow {\left({\log }_{2}x\right)}^2+4{\log }_{2}x-5=0$

Let $t={\log }_{2}x$

$\Rightarrow t^2+4t-5=0$

$\Rightarrow t^2+5t-t-5=0$

$\Rightarrow t(t+5)-1(t+5)=0$

$\Rightarrow (t+5)(t-1)=0$

$\therefore t=1,-5$

$\Rightarrow {\log }_{2}x=1,-5$

$\Rightarrow x=2^1,2^{-5}$

$\therefore x=2,\frac{1}{32}$