7.72 $\log _{\sqrt{3}} x \cdot \sqrt{\log _{\sqrt{3}} 3-\log _{x} 9}+4=0$.

Chapter 7. Logarithms. Exponential and Logarithmic Equations 221
7.72 For the sum on the left side of the equation to be zero, it is necessary that the first term be negative, i.e., $\log _{\sqrt{3}} x<0$. From this, it follows that

$0<x<1$.

Under this condition, $\log _{x} 9<0$ and, therefore, the expression under the square root is positive. Let's move 4 to the right side of the equation and square both sides:

$$
\log _{\sqrt{3}}^{2} x\left(2-2 \log _{x} 3\right)=16
$$

Since

$$
\log _{x} 3=\frac{1}{\log _{3} x}=\frac{1}{\log _{\sqrt{3}} \sqrt{x}}=\frac{2}{\log _{\sqrt{3}} x}
$$

then

$$
\log _{\sqrt{3}}^{2} x\left(2-\frac{4}{\log _{\sqrt{3}} x}\right)=16, \text { or } 2 \log _{\sqrt{3}}^{2} x-4 \log _{\sqrt{3}} x-16=0 \text {. }
$$

Assuming $\log _{\sqrt{3}} x=y<0$ (according to condition (1)) and solving the corresponding equation, we find $y_{1}=4, y_{2}=-2$. Only the value $y=-2$ satisfies the given equation, i.e., $\log _{\sqrt{3}} x=-2$, from which $x=\frac{1}{3}$.

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