Example 2 Solve the equation: $\log _{x} 2 \cdot \log _{2 x} 2=\log _{4 x} 2$.
Solve using the change of base formula to transform into
$$
\frac{1}{\log _{2} x} \cdot \frac{1}{\log _{2} 2 x}=\frac{1}{\log _{2} 4 x},
$$

and then into $\left(\log _{2} x\right)^{2}=2$, yielding
$$
\log _{2} x= \pm \sqrt{2},
$$

so $x=2^{\sqrt{2}}$ or $x=2^{-\sqrt{2}}$. Both are verified to be roots of the original equation.