3. Solve the inequality $\log _{5} 250+\left(4-\log _{5}^{2} 2\right) \log _{50} 5 \leqslant 125^{\log _{5}^{2} x}-24 \cdot x^{\log _{5} x}$. Answer. $x \in\left(0 ; \frac{1}{5}\right] \cup[5 ;+\infty)$.
Solution. Transform the left side of the inequality:

$\log _{5} 250+\left(4-\log _{5}^{2} 2\right) \log _{50} 5=3+\log _{5} 2+\frac{\left(2-\log _{5} 2\right)\left(2+\log _{5} 2\right)}{\log _{5} 50}=3+\log _{5} 2+\frac{\left(2-\log _{5} 2\right)\left(2+\log _{5} 2\right)}{2+\log _{5} 2}=5$

Let $5^{\log _{5}^{2} x}=v$. Then $x^{\log _{5} x}=\left(5^{\log _{5} x}\right)^{\log _{5} x}=v$ and the inequality becomes $5 \leqslant v^{3}-24 v \Leftrightarrow v^{3}-$ $24 v-5 \geqslant 0$.

One of the roots of the polynomial on the left side is $v=5$. Factoring out the factor $v-5$, we get $(v-$ 5) $\left(v^{2}+5 v+1\right) \geqslant 0$, from which $v \geqslant 5$ (since $v>0$, the second factor is positive). Find $x$ :

$$
5^{\log _{5}^{2} x} \geqslant 5 \Leftrightarrow \log _{5}^{2} x \geqslant 1 \Leftrightarrow\left[\begin{array}{l}
\log _{5} x \geqslant 1, \\
\log _{5} x \leqslant-1
\end{array} \Leftrightarrow x \in\left(0 ; \frac{1}{5}\right] \cup[5 ;+\infty)\right.
$$