3. Solve the inequality $\log _{9} 4+\left(16-\log _{3}^{2} 2\right) \log _{162} 3 \leqslant 64^{\log _{4}^{2} x}-15 \cdot x^{\log _{4} x}$.
Answer. $x \in\left(0 ; \frac{1}{4}\right] \cup[4 ;+\infty)$.

Solution. Transform the left side of the inequality:

$$
\log _{9} 4+\left(16-\log _{3}^{2} 2\right) \log _{162} 3=\log _{3} 2+\frac{\left(4-\log _{3} 2\right)\left(4+\log _{3} 2\right)}{\log _{3} 162}=\log _{3} 2+\frac{\left(4-\log _{3} 2\right)\left(4+\log _{3} 2\right)}{4+\log _{3} 2}=4
$$

Let $4^{\log _{4}^{2} x}=v$. Then $x^{\log _{4} x}=\left(4^{\log _{4} x}\right)^{\log _{4} x}=v$ and the inequality becomes $v^{3}-15 v-4 \geqslant 0$.

One of the roots of the polynomial on the left side is $v=4$. Factoring out the factor $v-4$, we get $(v-$ 4) $\left(v^{2}+4 v+1\right) \geqslant 0$, from which $v \geqslant 4$ (since $v>0$, the second factor is positive). We find $x$ :

$$
4^{\log _{4}^{2} x} \geqslant 4 \Leftrightarrow \log _{4}^{2} x \geqslant 1 \Leftrightarrow\left[\begin{array}{l}
\log _{4} x \geqslant 1 \\
\log _{4} x \leqslant-1
\end{array} \Leftrightarrow x \in\left(0 ; \frac{1}{4}\right] \cup[4 ;+\infty)\right.
$$