7.178. $\log _{5} x+\log _{25} x=\log _{1 / 5} \sqrt{3}$.
## Solution.

Domain of definition: $x>0$.

Let's switch to base 5. We have $\log _{5} x+\frac{1}{2} \log _{5} x=-\frac{1}{2} \log _{5} 3 \Leftrightarrow$
$\Leftrightarrow 2 \log _{5} x+\log _{5} x=\log _{5} \frac{1}{3} \Leftrightarrow \log _{5} x^{3}=\log _{5} \frac{1}{3}$. From this, we have $x^{3}=\frac{1}{3}$, $x=\frac{1}{\sqrt[3]{3}}$.

Answer: $\quad \frac{1}{\sqrt[3]{3}}$.