7.211. $\log _{\sqrt{3}} x \cdot \sqrt{\log _{\sqrt{3}} 3-\log _{x} 9}+4=0$.
Solution.

Domain of definition: $\left[\begin{array}{l}x \geq 3, \\ 0<x<1 .\end{array}\right.$

Let's transition to base 3. We get

$$
\begin{aligned}
& 2 \log _{3} x \cdot \sqrt{2-\frac{2}{\log _{3} x}}+4=0 \Leftrightarrow \log _{3} x \cdot \sqrt{\frac{2 \log _{3} x-2}{\log _{3} x}}=-2 \Rightarrow \\
& \Rightarrow\left\{\begin{array}{l}
\log _{3}^{2} x-\log _{3} x-2=0 \\
\log _{3} x<0
\end{array}\right.
\end{aligned}
$$

Solving this equation as a quadratic in terms of $\log _{3} x$, we have $\left(\log _{3} x\right)_{1}=-1$ or $\left(\log _{3} x\right)_{2}=2 ;\left(\log _{3} x\right)_{2}=2$ - an extraneous solution. Therefore, $x=3^{-1}=\frac{1}{3}$.

Answer: $\frac{1}{3}$.