7.181. $\log _{x} 3+\log _{3} x=\log _{\sqrt{x}} 3+\log _{3} \sqrt{x}+0.5$.
Solution.

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

Switch to base 3. We have

$\frac{1}{\log _{3} x}+\log _{3} x=\frac{2}{\log _{3} x}+\frac{1}{2} \log _{3} x+\frac{1}{2} \Leftrightarrow \log _{3}^{2} x-\log _{3} x-2=0 \Rightarrow$

$\Rightarrow\left(\log _{3} x\right)_{1}=-1$ or $\left(\log _{3} x\right)_{2}=2$, from which $x_{1}=\frac{1}{3}, x_{2}=9$.

Answer: $\quad \frac{1}{3} ; 9$.