7.195. $\log _{x}(125 x) \cdot \log _{25}^{2} x=1$.
## Solution.

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

Switch to base 5. Then we get $\frac{\log _{5} 125 x}{\log _{5} x} \cdot \frac{\log _{5}^{2} x}{\log _{5}^{2} 25}=1 \Leftrightarrow$
$\Leftrightarrow \log _{5}^{2} x+3 \log _{5} x-4=0$. Solving this equation as a quadratic equation in terms of $\log _{5} x$, we have $\left(\log _{5} x\right)_{1}=1$ or $\left(\log _{5} x\right)_{2}=-4$, from which $x_{1}=5$, $x_{2}=\frac{1}{625}$.

Answer: $\frac{1}{625} ; 5$.