1. Solve the inequality $\frac{\log _{3}\left(x^{4}\right) \cdot \log _{\frac{1}{3}}\left(x^{2}\right)+\log _{3}\left(x^{2}\right)-\log _{\frac{1}{3}}\left(x^{4}\right)+2}{\left(\log _{\frac{1}{3}}\left(x^{2}\right)\right)^{3}+64} \leq 0$.

“Phystech-2015”, 11th grade, solutions for ticket 5
Answer. $x \in(-9 ;-3] \cup\left[-\frac{1}{\sqrt[4]{3}} ; 0\right) \cup\left(0 ; \frac{1}{\sqrt[4]{3}}\right] \cup[3 ; 9)$.

Solution. The given inequality is equivalent to the following:

$$
\frac{2 \log _{3} x^{2} \cdot\left(-\log _{3} x^{2}\right)+\log _{3} x^{2}+2 \log _{3} x^{2}+2}{64-\log _{3}^{3} x^{2}} \leq 0
$$

After $\quad$ the substitution $\quad \log _{3} x^{2}=t \quad$ the inequality $\quad$ takes the form $\quad \frac{-2 t^{2}+3 t+2}{64-t^{3}} \leq 0, \quad$ from which $\quad \frac{(2-t)(1+2 t)}{4-t} \leq 0$, $t \in\left(-\infty ;-\frac{1}{2}\right] \cup[2 ; 4)$

We find the values of $x$.

For $t \leq-\frac{1}{2}$ we get $0<x^{2} \leq \frac{1}{\sqrt{3}} \Leftrightarrow x \in\left[-\frac{1}{\sqrt[4]{3}} ; 0\right) \cup\left(0 ; \frac{1}{\sqrt[4]{3}}\right]$.

For $2 \leq t<4$ we get $9 \leq x^{2}<81 \Leftrightarrow x \in(-9 ;-3] \cup[3 ; 9)$.