What is the value of log_3 7×log_5 9×log_7 11 ×log_9 13…log_21 25×log_23 27?

2018 AMC 12B Problem 7

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What is the value of

\begin{gathered} \log_3 7\cdot\log_5 9\cdot\log_7 11 \\ {}\cdot\log_9 13\cdots\log_{21} 25\cdot\log_{23} 27? \end{gathered}

3

3\log_7 23

6

9

10

Apply the change-of-base formula

\log_a b=\dfrac{\log b}{\log a}

Separate into two telescoping chains: bases

3,7,11,\dots

and bases

5,9,13,\dots

The factors split into two telescoping chains. The odd-position factors form

\begin{gathered} \log_3 7\cdot\log_7 11 \\ {}\cdot\log_{11} 15\cdots\log_{23} 27 \\ =\log_3 27=3, \end{gathered}

and the even-position factors form

\begin{gathered} \log_5 9\cdot\log_9 13\cdots\log_{21} 25 \\ =\log_5 25=2. \end{gathered}

The product is

3\cdot2=6.

Thus, the correct answer is C.