9.161. $0.4^{\log _{3} \frac{3}{x} \log _{3} 3 x}>6.25^{\log _{3} x^{2}+2}$.
## Solution.

Domain of definition: $x>0$.

Rewrite the inequality as $\left(\frac{2}{5}\right)^{\log _{3} \frac{3}{x} \log _{3} 3 x}>\left(\frac{2}{5}\right)^{-2 \log _{3} x^{2}-4} \Leftrightarrow$

$\Leftrightarrow \log _{3} \frac{3}{x} \log _{3} 3 x>-2 \log _{3} x^{2}-4 \Leftrightarrow$

$\Leftrightarrow\left(\log _{3} 3-\log _{3} x\right)\left(\log _{3} 3+\log _{3} x\right)0$.

Solving this inequality as a quadratic in terms of $\log _{3} x$, we get $\left[\begin{array}{l}\log _{3} x>5, \\ \log _{3} x<1.\end{array}\right.$

This implies $\left[\begin{array}{l}x>243, \\ 0<x<\frac{1}{3} .\end{array}\right.$

Answer: $x \in\left(0 ; \frac{1}{3}\right) \cup(243 ; \infty)$.