7.167. $\sqrt{\log _{x} \sqrt{5 x}}=-\log _{x} 5$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\log _{x} \sqrt{5 x} \geq 0, \\ -\log _{x} 5 \geq 0, \\ 0<x \neq 1\end{array}\right.$ or $0<x \leq \frac{1}{5}$.

By squaring both sides of the equation, we have or

$\log _{x} \sqrt{5 x}=\log _{x}^{2} 5 \Leftrightarrow 2 \log _{x}^{2} 5-\log _{x} 5-1=0 \Rightarrow\left(\log _{x} 5\right)_{h}=-\frac{1}{2}, x_{1}=\frac{1}{25}$

$\left(\log _{x} 5\right)=1, x_{2}=5$

$x_{2}=5$ does not satisfy the domain of definition.

Answer: $\frac{1}{25}$.