4・82 Solve the equation $\log _{3} x+\log _{x} 3-2 \log _{3} x \log _{x} 3=\frac{1}{2}$.
[Solution] The original equation is
$$\log _{3} x+\log _{x} 3-2=\frac{1}{2} .$$

Substituting $\log _{x} 3=\frac{1}{\log _{3} x}$ into the above equation and simplifying, we get
$$2 \log _{3}^{2} x-5 \log _{3} x+2=0$$

Thus, $\log _{3} x=\frac{1}{2}, \log _{3} x=2$.
This means $x=\sqrt{3}, x=9$.
Upon verification, both are solutions to the original equation.