AMC12 2018 B
AMC12 2018 B · Q7
AMC12 2018 B · Q7. It mainly tests Logarithms (rare).
What is the value of \log_3 7 \cdot \log_5 9 \cdot \log_7 11 \cdot \log_9 13 \cdot \ldots \cdot \log_{21} 25 \cdot \log_{23} 27?
\log_3 7 \cdot \log_5 9 \cdot \log_7 11 \cdot \log_9 13 \cdot \ldots \cdot \log_{21} 25 \cdot \log_{23} 27 的值为多少？
(A)
3
3
(B)
3 \log_7 23
3 \log_7 23
(C)
6
6
(D)
9
9
(E)
10
10
Answer
Correct choice: (C)
正确答案：（C）
Solution
Answer (C): The change of base formula states that $\log_a b=\frac{\log b}{\log a}$. Thus the product telescopes:
\[
\frac{\log 7}{\log 3}\cdot\frac{\log 9}{\log 5}\cdot\frac{\log 11}{\log 7}\cdot\frac{\log 13}{\log 9}\cdots\frac{\log 25}{\log 21}\cdot\frac{\log 27}{\log 23}
=\frac{\log 25}{\log 3}\cdot\frac{\log 27}{\log 5}
\]
\[
=\frac{\log 5^2}{\log 3}\cdot\frac{\log 3^3}{\log 5}
=\frac{2\log 5}{\log 3}\cdot\frac{3\log 3}{\log 5}
=6.
\]
答案（C）：换底公式表明 $\log_a b=\frac{\log b}{\log a}$。因此该乘积会望远镜式相消：
\[
\frac{\log 7}{\log 3}\cdot\frac{\log 9}{\log 5}\cdot\frac{\log 11}{\log 7}\cdot\frac{\log 13}{\log 9}\cdots\frac{\log 25}{\log 21}\cdot\frac{\log 27}{\log 23}
=\frac{\log 25}{\log 3}\cdot\frac{\log 27}{\log 5}
\]
\[
=\frac{\log 5^2}{\log 3}\cdot\frac{\log 3^3}{\log 5}
=\frac{2\log 5}{\log 3}\cdot\frac{3\log 3}{\log 5}
=6.
\]
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