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

2018 AMC 12B Problem 7 — Solution

2018 AMC 12B Problem 7

Below is the professionally curated solution for Problem 7 of the 2018 AMC 12B, 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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Concepts:logarithm telescoping

Difficulty rating: 1580

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

$3$

$3\log_7 23$

$6$

$9$

$10$

Solution:

The factors split into two telescoping chains. The odd-position factors form $\log_3 7\cdot\log_7 11\cdot\log_{11} 15\cdots\log_{23} 27=\log_3 27=3,$ and the even-position factors form $\log_5 9\cdot\log_9 13\cdots\log_{21} 25=\log_5 25=2.$

The product is $3\cdot2=6.$

Thus, the correct answer is C.

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