9.189. $\log _{5} \sqrt{3 x+4} \cdot \log _{x} 5>1$.
## Solution.

Domain of definition: $01$. From this, we have $\log _{x} \sqrt{3 x+4}>1$. The last inequality is equivalent to two systems of inequalities:

1) $\left\{\begin{array}{l}00,\end{array}\right.\right.\right.$
2); $\left\{\begin{array}{l}x>1, \\ \sqrt{3 x+4}>x\end{array} \Leftrightarrow\left\{\begin{array}{l}x>1, \\ 3 x+4>x^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}x>1, \\ (x-4)(x+1)<0\end{array} \Leftrightarrow 1<x<4\right.\right.\right.$.

Answer: $x \in(1 ; 4)$.