7.235. $x \log _{x+1} 5 \cdot \log _{\sqrt[3]{1 / 5}}(x+1)=\frac{x-4}{x}$.
## Solution.

$$
\text { Domain: }\left\{\begin{array}{l}
0<x+1 \neq 1, \\
x \neq 0
\end{array} \Leftrightarrow-1<x \neq 0\right.
$$

Switch to base 5. We have $\frac{x}{\log _{5}(x+1)} \cdot(-3) \log _{5}(x+1)=\frac{x-4}{x}$,

$-3 x=\frac{x-4}{x}$ when $\log _{5}(x+1) \neq 0$. From here, $3 x^{2}+x-4=0, x_{1}=-\frac{4}{3}, x_{2}=1$; $x_{1}=-\frac{4}{3}$ does not satisfy the domain.

Answer: 1.