9.076. $\sqrt{x^{\log _{2} \sqrt{x}}}>2$.

9.076. $\sqrt{x^{\log _{2} \sqrt{x}}}>2$.
## Solution.

Domain of definition: $x>0$.
From the condition $x^{\frac{\log _{2} \sqrt{x}}{2}}>2, x^{\frac{\log _{2} x}{4}}>2$. Taking the logarithm of both sides of the inequality with base 2, we get

$\log _{2} x^{\frac{\log _{2} x}{4}}>\log _{2} 2, \frac{\log _{2} x}{4} \cdot \log _{2} x>1, \log _{2}^{2} x>4 \Leftrightarrow\left|\log _{2} x\right|>2 \Rightarrow$

$\Rightarrow 1) \log _{2} x2$.

From this: 1) $04$.

Answer: $\quad x \in\left(0 ; \frac{1}{4}\right) \cup(4 ; \infty)$.