9.155. $0.2^{\frac{6 \log _{4} x-3}{\log _{4} x}}>\sqrt[3]{0.008^{2 \log _{4} x-1}}$.
Solution.

Let's write the equation as $0.2^{\frac{6 \log _{4} x-3}{\log _{4} x}}>0.2^{2 \log _{4} x-1} \Leftrightarrow$

$$
\begin{aligned}
& \Leftrightarrow \frac{6 \log _{4} x-3}{\log _{4} x}0 \Leftrightarrow \\
& \Leftrightarrow \frac{2\left(\log _{4} x-\frac{1}{2}\right)\left(\log _{4} x-3\right)}{\log _{4} x}>0 \Leftrightarrow\left(\log _{4} x-\frac{1}{2}\right)\left(\log _{4} x-3\right) \log _{4} x>0
\end{aligned}
$$

Using the interval method, we find

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0674.jpg?height=161&width=442&top_left_y=836&top_left_x=134)

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0674.jpg?height=132&width=558&top_left_y=1020&top_left_x=330)

Answer: $x \in(1 ; 2) \cup(64 ; \infty)$.