(9) If the real numbers $x, \alpha, \beta$ satisfy $x=\log _{3} \tan \alpha=-\log _{3} \tan \beta$, and $\alpha-\beta=$ $\frac{\pi}{6}$, then the value of $x$ is $\qquad$.
(9) $\frac{1}{2}$ Hint: Let $\tan \alpha=y$, then from the first condition we get $\tan \beta=\frac{1}{y}$. Also, from the second condition, we know
$$
\tan \beta=\tan \left(\alpha-\frac{\pi}{6}\right)=\frac{y-\frac{1}{\sqrt{3}}}{1+\frac{y}{\sqrt{3}}},
$$

thus we obtain the equation
$$
\frac{1}{y}=\frac{y-\frac{1}{\sqrt{3}}}{1+\frac{y}{\sqrt{3}}}
$$

Taking the positive root, we get $y=\sqrt{3}$. Therefore, $x=\log _{3} y=\frac{1}{2}$.