Question

Solve the following equation:
$3\sqrt{{\log }_{2}x}-{\log }_{2}8x+1=0$

Answer 1

$3\sqrt{{\log }_{2}x}-{\log }_{2}8x+1=0\Rightarrow 16+{\log }_{2}x=a\Rightarrow 3\sqrt{a}-(3+a)+1=0\Rightarrow -a+3\sqrt{a}-2=0\Rightarrow \left(\sqrt{a-1)}\right(\sqrt{a-2})=0\Rightarrow \sqrt{a}=1,\sqrt{a}=2\Rightarrow a=1,a=4\Rightarrow x=2^1,2^4\Rightarrow x=2,16.$