If $x^{\sqrt{x}}=\sqrt{x^x},$ then $x=$

$x^{\sqrt{x}}=\sqrt{x^x}$

$$\begin{gathered} x^{\sqrt{x}}={\left(x^x\right)}^{1/2} \\ {x}^{\sqrt{x}} = x^{\frac{x}{2}} \\ \therefore \sqrt{x} = \frac{x}{2} \\ squaring on both sides \\ x= \frac{x^2}{4} \\ 4x = x^2 \\ \frac{x^2}{x} = 4 \\ x = 4 \end{gathered}$$