$6 \cdot 38$ If $y=\log _{5} 6 \cdot \log _{6} 7 \cdot \log _{7} 8 \cdot \log _{8} 9 \cdot \log _{9} 10$, then
(A) $y \in(0,1)$.
(B) $y=1$.
(C) $y \in(1,2)$.
(D) $y=2$.
(E) $y \in(2,3)$.
(Shanghai High School Mathematics Competition, 1982)
［Solution］Using the change of base formula for logarithms, we have
$$
\begin{array}{l}
\quad y=\frac{\lg 6}{\lg 5} \cdot \frac{\lg 7}{\lg 6} \cdot \frac{\lg 8}{\lg 7} \cdot \frac{\lg 9}{\lg 8} \cdot \frac{\lg 10}{\lg 9}=\frac{\lg 10}{\lg 5}=\log _{5} 10 . \\
\because \quad \log _{5} 5<\log 5<\log _{5} 25,
\end{array}
$$

Thus, $1<y<2$.
Therefore, the answer is $(C)$.