2. Solve the inequality $x^{\log _{3} x}-2 \leqslant(\sqrt[3]{3})^{\log _{\sqrt{3}}^{2} x}-2 \cdot x^{\log _{3}} \sqrt[3]{x}$.
Answer. $x \in\left(0 ; 3^{-\sqrt{\log _{3} 2}}\right] \cup\{1\} \cup\left[3 \sqrt{\log _{3} 2} ;+\infty\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_543143f47b43f01a28cfg-01.jpg?height=60&width=1819&top_left_y=1272&top_left_x=153)
Matching the third with the second, and the second with the fourth, we factorize the left side of the inequality:

There are two possible cases.

$$
\left(3^{\log _{3}^{2} x}-2\right)\left(1-3^{\frac{1}{3} \log _{3}^{2} x}\right) \leqslant 0
$$

a) $\left\{\begin{array}{l}3^{\log _{3}^{2} x}-2 \geqslant 0, \\ 1-3^{\frac{1}{3} \log _{3}^{2} x} \leqslant 0\end{array} \Leftrightarrow\left\{\begin{array}{l}\log _{3}^{2} x \geqslant \log _{3} 2, \\ \log _{3}^{2} x \geqslant 0\end{array} \Leftrightarrow\left[\begin{array}{l}\log _{3} x \geqslant \sqrt{\log _{3} 2}, \\ \log _{3} x \leqslant-\sqrt{\log _{3} 2}\end{array} \Leftrightarrow\left[\begin{array}{l}x \geqslant 3^{\sqrt{\log _{3} 2}}, \\ x \leqslant 3^{-\sqrt{\log _{3} 2}} ;\end{array}\right.\right.\right.\right.$

b) $\left\{\begin{array}{l}3^{\log _{3}^{2} x}-2 \leqslant 0, \\ 1-3^{\frac{1}{3} \log _{3}^{2} x} \geqslant 0\end{array} \Leftrightarrow\left\{\begin{array}{l}\log _{3}^{2} x \leqslant \log _{3} 2, \\ \log _{3}^{2} x \leqslant 0\end{array} \Leftrightarrow \log _{3} x=0 \Leftrightarrow x=1\right.\right.$.

Combining the results, we get $x \in\left(0 ; 3^{-\sqrt{\log _{3} 2}}\right] \cup\{1\} \cup\left[3^{\sqrt{\log _{3} 2}} ;+\infty\right)$.