7.241. $\frac{\log _{2}\left(x^{3}+3 x^{2}+2 x-1\right)}{\log _{2}\left(x^{3}+2 x^{2}-3 x+5\right)}=\log _{2 x} x+\log _{2 x} 2$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}x^{3}+3 x^{2}+2 x-1>0, \\ 0<x^{3}+2 x^{2}-3 x+5 \neq 1, \\ 0<x \neq \frac{1}{2} .\end{array}\right.$

By the formula for changing the base we have

$$
\begin{aligned}
& \log _{x^{3}+2 x^{2}-3 x+5}\left(x^{3}+3 x^{2}+2 x-1\right)=\log _{2 x} 2 x \Leftrightarrow \\
& \Leftrightarrow \log _{x^{3}+2 x^{2}-3 x+5}\left(x^{3}+3 x^{2}+2 x-1\right)=1 \Leftrightarrow
\end{aligned}
$$

$\Leftrightarrow x^{3}+3 x^{2}+2 x-1=x^{3}+2 x^{2}-3 x+5 \Leftrightarrow x^{2}+5 x-6=0 \Rightarrow x_{1}=1$, $x_{2}=-6 ; x_{2}=-6$ does not satisfy the domain of definition.

Answer: 1 .