$4 \cdot 87$ Solve the equation $\log _{x} 3 \cdot \log _{3 x} 3=\log _{9 x} 3$.
[Solution] By the definition of logarithm, we know that $x>0$, and $x \neq 1, x \neq \frac{1}{3}, x \neq \frac{1}{9}$. Using the change of base formula, the original equation can be rewritten as:
$$
\frac{1}{\log _{3} x} \cdot \frac{1}{\log _{3} 3 x}=\frac{1}{\log _{3} 9 x},
$$

which is
$$
\begin{aligned}
\log _{3} 9 x & =\log _{3} x \cdot \log _{3} 3 x, \\
2+\log _{3} x & =\log _{3} x\left(\log _{3} x+1\right),
\end{aligned}
$$

which simplifies to $\left(\log _{3} x\right)^{2}=2$, hence $\log _{3} x= \pm \sqrt{2}$,
so $\quad x=3^{ \pm \sqrt{2}}$.