Question

Solve the following equation:
$\sqrt{{\log }_{2}x}-0.5={\log }_2\sqrt{x}$

Answer 1

$\sqrt{{\log }_{2}x}-0.5={\log }_2\sqrt{x}\Rightarrow 2\sqrt{{\log }_{2}x}-1={\log }_{2}x\Rightarrow {\left(\sqrt{{\log }_{2}x}\right)}^2-2\sqrt{{\log }_{2}x}+1=0\Rightarrow \sqrt{{\log }_{2}x}=1\Rightarrow {\log }_{2}x=1\Rightarrow x=2.$