7.081. $\log _{5} x+\log _{x} 25=\operatorname{ctg}^{2} \frac{25 \pi}{6}$.
## Solution.

Domain of definition: $0<x \neq 1$.

Switch to base 5. We have

$$
\begin{aligned}
& \log _{5} x+\frac{2}{\log _{5} x}=\left(\operatorname{ctg}\left(4 \pi+\frac{\pi}{6}\right)\right)^{2}, \log _{5} x+\frac{2}{\log _{5} x}=3 \Rightarrow \\
& \Rightarrow \log _{5}^{2} x-3 \log _{5} x+2=0
\end{aligned}
$$

Solving this equation as a quadratic in $\log _{5} x$, we get $\left(\log _{5} x\right)_{1}=1$ or $\left(\log _{5} x\right)_{2}=2$, hence $x_{1}=5 ; x_{2}=25$.

Answer: $5 ; 25$.