$8 \cdot 27$ If $b>1, x>0$, and $(2 x)^{\log _{b} 2}-(3 x)^{\log _{b} 3}=0$, then $x$ is
(A) $\frac{1}{216}$.
(B) $\frac{1}{6}$.
(C) 1 .
(D) 6 .
(E) cannot be uniquely determined.
(32nd American High School Mathematics Examination, 1981)
[Solution] The given equation is equivalent to the following equation:
$$
(2 x)^{\log _{b} 2}=(3 x)^{\log _{b} 3} \text {, }
$$

or $\frac{2^{\log _{b} 2}}{3^{\log _{b} 3}}=\frac{x^{\log _{b} 3}}{x^{\log _{b} 2}}$, which simplifies to $\frac{2^{\log _{b} 2}}{3^{\log _{b} 3}}=x^{\log _{b} 3-\log _{b} 2}$.
Taking the logarithm with base $b$ on both sides of the above equation, we get
$$
\begin{array}{l} 
\log _{b} 2^{\log _{b} 2}-\log _{b} 3^{\log _{b} 3}=\left(\log _{b} 3-\log _{b} 2\right) \log _{b} x, \\
\left(\log _{b} 2\right)^{2}-\left(\log _{b} 3\right)^{2}=\left(\log _{b} 3-\log _{b} 2\right) \log _{b} x .
\end{array}
$$

Thus,
$$
-\left(\log _{b} 2+\log _{b} 3\right)=\log _{b} x,
$$

or
$$
-\log _{b}(2 \cdot 3)=\log _{b} x,
$$

Therefore, $\log _{b} \frac{1}{6}=\log _{b} x$, which gives $x=\frac{1}{6}$.
Hence, the answer is $(B)$.