7.194. $\log _{x} 2-\log _{4} x+\frac{7}{6}=0$.
## Solution.

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

Switch to base 2. We have $\frac{1}{\log _{2} x}-\frac{1}{2} \log _{2} x+\frac{7}{6}=0 \Leftrightarrow$ $\Leftrightarrow 3 \log _{2}^{2} x-7 \log _{2} x-6=0$. Solving this equation as a quadratic in terms of $\log _{2} x$, we find $\left(\log _{2} x\right)_{1}=-\frac{2}{3}$ or $\left(\log _{2} x\right)_{2}=3$, from which $x_{1}=\frac{1}{\sqrt[3]{4}}, x_{2}=8$.

Answer: $\quad \frac{1}{\sqrt[3]{4}} ; 8$.