7.095. $\sqrt{\log _{a} x}+\sqrt{\log _{x} a}=\frac{10}{3}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\log _{a} x \geq 0, \\ 0<a \neq 1, \\ 0<x \neq 1 .\end{array}\right.$

From the condition

$$
\sqrt{\log _{a} x}+\frac{1}{\sqrt{\log _{a} x}}-\frac{10}{3}=0 \Rightarrow 3\left(\sqrt{\log _{a} x}\right)^{2}-10 \sqrt{\log _{a} x}+3=0
$$

Solving this equation as a quadratic equation in terms of $\sqrt{\log _{a} x}$, we get $\left(\sqrt{\log _{a} x}\right)_{1}=\frac{1}{3},\left(\log _{a} x\right)_{1}=\frac{1}{9}$, from which $x_{1}=\sqrt[9]{a}$, or $\left(\sqrt{\log _{a} x}\right)_{2}=3$, $\left(\log _{a} x\right)_{2}=9$, from which $x_{2}=a^{9}$.

Answer: $\sqrt[9]{a} ; a^{9}$, where $0<a \neq 1$.