9.178. $x^{0.5 \log _{0.5} x-3} \geq 0.5^{3-2.5 \log _{0.5} x}$.
## Solution.

Domain of definition: $x>0$.

By logarithmizing both sides of the inequality with base 0.5, we get

$$
\begin{aligned}
& \log _{0.5} x^{0.5 \log _{0.5} x-3} \leq \log _{0.5} 0.5^{3-2.5 \log _{0.5} x} \Leftrightarrow \\
& \Leftrightarrow\left(0.5 \log _{0.5} x-3\right) \log _{0.5} x \leq\left(3-2.5 \log _{0.5} x\right) \log _{0.5} 0.5 \Leftrightarrow \\
& \Leftrightarrow 0.5 \log _{0.5}^{2} x-3 \log _{0.5} x \leq 3-2.5 \log _{0.5} x \Leftrightarrow \\
& \Leftrightarrow 0.5 \log _{0.5}^{2} x-0.5 \log _{0.5} x-3 \leq 0
\end{aligned}
$$

Solving this inequality as a quadratic equation in terms of $\log _{0.5} x$, we find $-2 \leq \log _{0.5} x \leq 3 \Leftrightarrow 0.125 \leq x \leq 4$.

Answer: $x \in [0.125 ; 4]$.