7.106. $\log _{x} \sqrt{2}-\log _{x}^{2} \sqrt{2}=\log _{3} 27-\log _{x}(2 x)$.
Solution.

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

Rewrite the equation as

$$
\frac{1}{2} \log _{x} 2-\frac{1}{4} \log _{x}^{2} 2=3-\log _{x} 2-1, \quad \log _{x}^{2} 2-6 \log _{x} 2+8=0
$$

Solving this equation as a quadratic in terms of $\log _{x} 2$, we find $\log _{x} 2=2, \log _{x} 2=4$, hence $x^{2}=2$ or $x^{4}=2$. Therefore, $x_{1}=-\sqrt{2}$, $x_{2}=\sqrt{2}, x_{3}=-\sqrt[4]{2}, x_{4}=\sqrt[4]{2} ; x_{1}=-\sqrt{2}$ and $x_{3}=-\sqrt[4]{2}$ do not satisfy the domain of definition.

Answer: $\sqrt[4]{2} ; \sqrt{2}$.