2. Solve the inequality $(\sqrt[7]{4})^{\log _{\sqrt{2}}^{2} x}+6 \geqslant x^{\log _{2} x}+6(\sqrt[7]{x})^{\log _{2} x}$.
Answer. $x \in\left(0 ; 2^{-\sqrt{\log _{2} 6}}\right] \cup\{1\} \cup\left[2 \sqrt{\log _{2} 6} ;+\infty\right)$.

Solution. Rewrite the inequality as $2^{\frac{8}{7} \log _{2}^{2} x}+6-2^{\log _{2}^{2} x}-6 \cdot 2^{\frac{1}{7} \log _{2}^{2} x} \geqslant 0$. Grouping the first term with the third and the second with the fourth, we factor the left side of the inequality:

$$
\left(2^{\log _{2}^{2} x}-6\right)\left(2^{\frac{1}{7} \log _{2}^{2} x}-1\right) \geqslant 0
$$

There are two possible cases.

a) $\left\{\begin{array}{l}2^{\log _{2}^{2} x}-6 \geqslant 0, \\ 2^{\frac{1}{7} \log _{2}^{2} x}-1 \geqslant 0\end{array} \Leftrightarrow\left\{\begin{array}{l}\log _{2}^{2} x \geqslant \log _{2} 6, \\ \log _{2}^{2} x \geqslant 0\end{array} \Leftrightarrow\left[\begin{array}{l}\log _{2} x \geqslant \sqrt{\log _{2} 6}, \\ \log _{2} x \leqslant-\sqrt{\log _{2} 6}\end{array} \Leftrightarrow\left[\begin{array}{l}x \geqslant 2^{\sqrt{\log _{2} 6}}, \\ x \leqslant 2^{-\sqrt{\log _{2} 6}} ;\end{array}\right.\right.\right.\right.$

![](https://cdn.mathpix.com/cropped/2024_05_06_543143f47b43f01a28cfg-10.jpg?height=131&width=1063&top_left_y=1699&top_left_x=151)

Combining the results, we get $x \in\left(0 ; 2^{-\sqrt{\log _{2} 6}}\right] \cup\{1\} \cup\left[2^{\sqrt{\log _{2} 6}} ;+\infty\right)$.