9.165. $2^{\log _{0.5}^{2} x}+x^{\log _{0.5} x}>2.5$.
## Solution.

Domain of definition: $x>0$.

Rewrite the inequality as $2^{\log _{0.5} x \cdot \log _{0.5} x}+x^{\log _{0.5} x}-2.5>0 \Leftrightarrow$ $\Leftrightarrow\left(2^{\log _{0.5} x}\right)^{\log _{0.5} x}+x^{\log _{0.5} x}-2.5>0 \Leftrightarrow\left(2^{\log _{2} \frac{1}{x}}\right)^{\log _{0.5} x}+x^{\log _{0.5} x}-2.5>0 \Leftrightarrow$
$\Leftrightarrow \frac{1}{x^{\log _{0.5} x}}+x^{\log _{0.5} x}-2.5>0$. Since $00$. Solving this inequality as a quadratic

in terms of $x^{\log _{0.5} x}$, we get $\left[\begin{array}{l}x^{\log _{0.5} x}>2, \\ x^{\log _{0.5} x} \log _{2} 2, } \\
{ \log _{2} x^{\log _{0.5} x} 1, } \\
{ \log _{0.5} x \log _{2} x 1, \\
-\log _{2}^{2} x1
\end{array} \Leftrightarrow\left(\log _{2} x-1\right)\left(\log _{2} x+1\right)>0 \Leftrightarrow \\
& \Leftrightarrow\left[\begin{array} { l } 
{ \log _{2} x > 1, } \\
{ \log _{2} x 2, \\
0<x<\frac{1}{2} .
\end{array}\right.\right. \\
& \text { Answer: } x \in\left(0 ; \frac{1}{2}\right) \cup(2 ; \infty) .
\end{aligned}
$$