3. Solve the inequality $27^{\log _{3}^{2} x}-8 \cdot x^{\log _{3} x} \geqslant \log _{25} 4+\left(9-\log _{5}^{2} 2\right) \log _{250} 5$.
Answer. $x \in\left(0 ; \frac{1}{3}\right] \cup[3 ;+\infty)$.

Solution. Transform the right side of the inequality:

$$
\log _{25} 4+\left(9-\log _{5}^{2} 2\right) \log _{250} 5=\log _{5} 2+\frac{\left(3-\log _{5} 2\right)\left(3+\log _{5} 2\right)}{\log _{5} 250}=\log _{5} 2+\frac{\left(3-\log _{5} 2\right)\left(3+\log _{5} 2\right)}{3+\log _{5} 2}=3
$$

Let $3^{\log _{3}^{2} x}=v$. Then $x^{\log _{3} x}=\left(3^{\log _{3} x}\right)^{\log _{3} x}=v$ and the inequality becomes $v^{3}-8 v-3 \geqslant 0$.

One of the roots of the polynomial on the left side is $v=3$. Factoring out the factor $v-3$, we get $(v-3)\left(v^{2}+3 v+1\right) \geqslant 0$, from which $v \geqslant 3$ (since $v>0$, the second factor is positive). We find $x$:

$$
3^{\log _{3}^{2} x} \geqslant 3 \Leftrightarrow \log _{3}^{2} x \geqslant 1 \Leftrightarrow\left[\begin{array}{l}
\log _{3} x \geqslant 1 \\
\log _{3} x \leqslant-1
\end{array} \Leftrightarrow x \in\left(0 ; \frac{1}{3}\right] \cup[3 ;+\infty)\right.
$$