What is the value of \begingathered \log_3 7×\log_5 9×\log_7 11 \ ×\log_9 13\cdots\log_21 25×\log_23 27? \endgathered

2018 AMC 12B Problem 7 — Solution

2018 AMC 12B Problem 7

Attempt Problem 7 of the 2018 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12B solutions, or check the answer key.

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What is the value of $\begin{gathered} \log_3 7\cdot\log_5 9\cdot\log_7 11 \\ {}\cdot\log_9 13\cdots\log_{21} 25\cdot\log_{23} 27? \end{gathered}$

$3$

$3\log_7 23$

$6$

$9$

$10$

Answer: C

Concepts:logarithm telescoping

Difficulty rating: 1580

Solution:

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.

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