4. Solve the inequality

$$
\log _{5}\left(5 x^{2}+2 x\right) \cdot \log _{5}\left(5+\frac{2}{x}\right)>\log _{5} 5 x^{2}
$$
Answer: $\left(\frac{-1-\sqrt{2}}{5} ;-\frac{2}{5}\right) \cup\left(\frac{-1+\sqrt{2}}{5} ; \infty\right)$.

Solution.

$$
\begin{aligned}
\log _{5}\left(5+\frac{2}{x}\right) \cdot \log _{5}\left(5 x^{2}+2 x\right)>\log _{5} 5 x^{2} & \Leftrightarrow \log _{5}\left(5+\frac{2}{x}\right)\left(\log _{5}\left(5+\frac{2}{x}\right)+\log _{5} x^{2}\right)>1+\log _{5} x^{2} \Leftrightarrow \\
& \Leftrightarrow\left(\log _{5}\left(5+\frac{2}{x}\right)-1\right)\left(\log _{5}\left(5+\frac{2}{x}\right)+\log _{5} x^{2}+1\right)>0
\end{aligned}
$$

since $5+\frac{2}{x}>0 \Leftrightarrow x \in\left(-\infty ;-\frac{2}{5}\right) \cup(0 ; \infty)$, we have two cases:

1) $x<0\left(\Rightarrow \log _{5}\left(5+\frac{2}{x}\right)>1\right): \log _{5}\left(5 x^{2}+2 x\right)>-1 \Leftrightarrow 25 x^{2}+10 x-1>0 \Leftrightarrow x \in\left(x_{+} ; \infty\right)$.