7.213. $\left|\log _{\sqrt{3}} x-2\right|-\left|\log _{3} x-2\right|=2$.
## Solution.

Domain of definition: $x>0$.

Switch to base 3. Then $\left|2 \log _{3} x-2\right|-\left|\log _{3} x-2\right|=2$. By opening the absolute values, we get three cases:

1) $\left\{\begin{array}{l}\log _{3} x<1, \\ -2 \log _{3} x+2+\log _{3} x-2=2\end{array} \Leftrightarrow\left\{\begin{array}{l}\log _{3} x<1, \\ \log _{3} x=-2\end{array} \Rightarrow x_{1}=3^{-2}=\frac{1}{9}\right.\right.$;
2) $\left\{\begin{array}{l}1 \leq \log _{3} x<2, \\ 2 \log _{3} x-2+\log _{3} x-2=2\end{array} \Leftrightarrow\left\{\begin{array}{l}1 \leq \log _{3} x<2, \\ \log _{3} x=2 .\end{array}\right.\right.$

$\log _{3} x=2$ does not fit because $\log _{3} x<2$.
3) $\left\{\begin{array}{l}\log _{3} x \geq 2, \\ 2 \log _{3} x-2-\log _{3} x+2=2\end{array} \Leftrightarrow\left\{\begin{array}{l}\log _{3} x \geq 2, \\ \log _{3} x=2\end{array} \Rightarrow x_{2}=3^{2}=9\right.\right.$.

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