9.175. $\log _{\frac{1}{5}} x+\log _{4} x>1$.
## Solution.

Domain of definition: $x>0$.

Switch to base 4. We have $\frac{\log _{4} x}{\log _{4} \frac{1}{5}}+\log _{4} x>1 \Leftrightarrow$

$\Leftrightarrow \frac{\log _{4} x}{-\log _{4} 5}+\log _{4} x>1 \Leftrightarrow \log _{4} x \cdot\left(\frac{1-\log _{4} 5}{-\log _{4} 5}\right)>1 \Leftrightarrow$

$\Leftrightarrow \log _{4} x \cdot\left(\frac{\log _{4} 5-1}{\log _{4} 5}\right)>1 \Leftrightarrow \log _{4} x>\frac{\log _{4} 5}{\log _{4} 5-1}, \log _{4} x>\frac{\log _{4} 5}{\log _{4} \frac{5}{4}} \Leftrightarrow$

$\Leftrightarrow \log _{4} x>\log _{\frac{5}{4}} 5 \Leftrightarrow x>4^{\log _{\frac{5}{4}} 5}$.

Further, $4^{\log _{\frac{5}{5}} 5}=4^{\log _{\frac{5}{4}}\left(\frac{1}{5}\right)^{-1}}=4^{-\log _{\frac{5}{4}}^{4} 0.2}=4^{\log _{\frac{4}{5}} 0.2}=4^{\log _{0.8} 0.2}$.

Answer: $x \in\left(4^{\log _{0.8} 0.2} ; \infty\right)$.