$6 \cdot 39$ If $y=\left(\log _{2} 3\right)\left(\log _{3} 4\right) \cdot \cdots \cdot\left(\log _{n}[n+1]\right) \cdot \cdots \cdot$ $\left(\log _{31} 32\right)$, then
(A) $4<y<5$.
(B) $y=5$.
(C) $5<y<6$.
(D) $y=6$.
(E) $6<y<7$.
(28th American High School Mathematics Examination, 1977)
［Solution］ By the change of base formula, we have
$$
\begin{aligned}
y & =\frac{\lg 3}{\lg 2} \cdot \frac{\lg 4}{\lg 3} \ldots \cdot \frac{\lg (n+1)}{\lg n} \ldots \cdot \frac{\lg 31}{\lg 30} \cdot \frac{\lg 32}{\lg 31} \\
& =\frac{\lg 32}{\lg 2}=\log _{2} 32=5 .
\end{aligned}
$$

Therefore, the correct choice is $(B)$.