5. Solve the inequality $\log _{1+x^{2}}\left(1-8 x^{5}\right)+\log _{1-x^{2}+8 x^{4}}\left(1+x^{2}\right) \leqslant 1+\log _{1-x^{2}+8 x^{4}}\left(1-8 x^{5}\right)$.
Answer: $x \in\left\{-\frac{1}{2}\right\} \bigcup\left(-\frac{1}{2 \sqrt{2}} ; 0\right) \cup\left(0 ; \frac{1}{2 \sqrt{2}}\right) \cup\left[\frac{1}{2} ; \frac{1}{\sqrt[5]{8}}\right)$.

Solution. Let $1+x^{2}=u, 1-x^{2}+8 x^{4}=v, 1-8 x^{5}=w$. Then the inequality takes the form $\log _{u} w+\log _{v} u-\log _{v} w-1 \leqslant 0$. This can be transformed as follows:

$$
\begin{aligned}
& \log _{u} w+\frac{1}{\log _{u} v}-\frac{\log _{u} w}{\log _{u} v}-1 \leqslant 0 \Leftrightarrow \frac{\log _{u} v \log _{u} w+1-\log _{u} w-\log _{u} v}{\log _{u} v} \leqslant 0 \Leftrightarrow \\
& \Leftrightarrow \frac{\left(\log _{u} w-1\right)\left(\log _{u} v-1\right)}{\log _{u} v} \leqslant 0 \Leftrightarrow \frac{\left(\log _{u} w-\log _{u} u\right)\left(\log _{u} v-\log _{u} u\right)}{\log _{u} v} \leqslant 0
\end{aligned}
$$

To solve this inequality, we further apply the method of rationalization: the sign of the difference $\log _{a} b- \log _{a} c$ on the domain of valid values coincides with the sign of the expression $\frac{b-c}{a-1}$; in particular (when $c=1$), the sign of the logarithm $\log _{a} b$ coincides with the sign of the expression $\frac{b-1}{a-1}$. Then from the last inequality we get

$$
\frac{(w-u)(v-u)}{(u-1)(v-1)} \leqslant 0
$$

The domain of the original inequality is defined by the conditions $u>0, v>0, w>0, u \neq 1, v \neq 1$. The last two restrictions are automatically satisfied for any solution (??), since when $u=1$ or $v=1$ the denominator of the fraction becomes zero. In addition, $u=1+x^{2}$ and $v=1-x^{2}+8 x^{4}$ are positive for all values of $x$. Therefore, the only restriction from the domain that needs to be considered is the inequality $1-8 x^{5}>0$, from which $x<\frac{1}{\sqrt[5]{8}}$. We solve the inequality $(? ?):$

$$
\begin{aligned}
\frac{\left(1-8 x^{5}-x^{2}-1\right)\left(1-x^{2}+8 x^{4}-1-x^{2}\right)}{\left(1+x^{2}-1\right)\left(1-x^{2}+8 x^{4}-1\right)} \leqslant 0 \Leftrightarrow & \Leftrightarrow \frac{2 x^{4}\left(-8 x^{3}-1\right)\left(4 x^{2}-1\right)}{x^{4}\left(8 x^{2}-1\right)} \leqslant 0 \Leftrightarrow \\
\Leftrightarrow\left\{\begin{array}{l}
x \neq 0, \\
\frac{(2 x+1)\left(4 x^{2}-2 x+1\right)(2 x+1)(2 x-1)}{\left(8 x^{2}-1\right)} \geqslant 0
\end{array}\right. & \Leftrightarrow\left\{\begin{array}{l}
x \neq 0, \\
\frac{(2 x-1)(2 x+1)^{2}}{\left(x-\frac{1}{2 \sqrt{2}}\right)\left(x+\frac{1}{2 \sqrt{2}}\right)} \geqslant 0
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x \neq 0, \\
x \in\left\{-\frac{1}{2}\right\} \cup\left(-\frac{1}{2 \sqrt{2}} ; \frac{1}{2 \sqrt{2}}\right) \cup\left[\frac{1}{2} ;+\infty\right)
\end{array}\right.
\end{aligned}
$$

Considering the domain, we finally get $x \in\left\{-\frac{1}{2}\right\} \bigcup\left(-\frac{1}{2 \sqrt{2}} ; 0\right) \cup\left(0 ; \frac{1}{2 \sqrt{2}}\right) \cup\left[\frac{1}{2} ; \frac{1}{\sqrt[5]{8}}\right)$.