4・87 Solve the equation $\log _{x} 3 \cdot \log _{3 x} 3=\log _{9} 3$.


[Solution] By the definition of logarithms, 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:

i.e., $\square$
$$\frac{1}{\log _{3} x} \cdot \frac{1}{\log _{3} 3 x}=\frac{1}{\log _{3} 9 x}$$
$$\begin{array}{l}
\log _{3} 9 x=\log _{3} x \cdot \log _{3} 3 x \\
2+\log _{3} x=\log _{3} x\left(\log _{3} x+1\right)
\end{array}$$

i.e., $\square$
$$\left(\log _{3} x\right)^{2}=2 \text {, hence } \log _{3} x= \pm \sqrt{2} \text {, }$$

Therefore, $\quad x=3^{ \pm \sqrt{2}}$.