Question

If $\sqrt{{\log }_{2}x}-0.5={\log }_2\sqrt{x}$, then $x$ equals to

Answer 1

$\sqrt{{\log }_{2}x}-0.5={\log }_2\sqrt{x}$
For the equation to be defined,
${\log }_{2}x\ge 0$ and $\sqrt{x}>0$
$\Rightarrow x\ge 1$ and $x>0$
$\Rightarrow x\ge 1$

Now, $\sqrt{{\log }_{2}x}-0.5={\log }_2\sqrt{x}$
$\Rightarrow \sqrt{{\log }_{2}x}-\frac{1}{2}=\frac{1}{2}{\log }_{2}x$
Let ${\log }_{2}x=t\Rightarrow t\ge 0$
Then $\sqrt{t}-\frac{1}{2}=\frac{1}{2}t$
$\Rightarrow 2\sqrt{t}-1=t\Rightarrow (\sqrt{t}{)}^2-2\sqrt{t}+1=0\Rightarrow \sqrt{t}=1\Rightarrow t=1\therefore x=2$