9.159. $2 \log _{3} \log _{3} x+\log _{\frac{1}{3}} \log _{3}(9 \sqrt[3]{x}) \geq 1$.
## Solution.

Domain: $\log _{3} x>0$.

Switch to base 3. $2 \log _{3} \log _{3} x-\log _{3} \log _{3} 9 \sqrt[3]{x} \geq 1 \Leftrightarrow$ $\Leftrightarrow \log _{3} \log _{3}^{2} x-\log _{3} \log _{3} 9 \sqrt[3]{x} \geq 1, \log _{3} \frac{\log _{3}^{2} x}{\log _{3} 9 \sqrt[3]{x}} \geq 1, \frac{\log _{3}^{2} x}{\log _{3} 9 \sqrt[3]{x}} \geq 3 \Leftrightarrow$

$$
\begin{aligned}
& \Leftrightarrow \frac{\log _{3}^{2} x}{2+\frac{1}{3} \log _{3} x}-3 \geq 0 \Leftrightarrow \frac{\log _{3}^{2} x-\log _{3} x-6}{\log _{3} x+6} \geq 0 \\
& \frac{\left(\log _{3} x+2\right)\left(\log _{3} x-3\right)}{\log _{3} x+6} \geq 0 \Leftrightarrow\left\{\begin{array}{l}
\left(\log _{3} x+2\right)\left(\log _{3} x-3\right)\left(\log _{3} x+6\right) \geq 0 \\
\log _{3} x+6 \neq 0
\end{array}\right.
\end{aligned}
$$

With respect to $\log _{3} x$ using the interval method, we get

1) $-6<\log _{3} x \leq-2$, does not fit the domain;
2) $\log _{3} x \geq 3 \Leftrightarrow x \geq 27$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0677.jpg?height=88&width=505&top_left_y=608&top_left_x=353)

Answer: $x \in[27 ; \infty)$.