2015 AMC 12A Problem 14

Below is the professionally curated solution for Problem 14 of the 2015 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 12A solutions, or check the answer key.

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Concepts:logarithm

Difficulty rating: 1730

14.

What is the value of aa for which 1log2a+1log3a+1log4a=1?\dfrac{1}{\log_2 a} + \dfrac{1}{\log_3 a} + \dfrac{1}{\log_4 a} = 1?

99

1212

1818

2424

3636

Solution:

By the change-of-base formula, 1logba=logab.\dfrac{1}{\log_b a} = \log_a b. Therefore 1=loga2+loga3+loga4=loga24.1 = \log_a 2 + \log_a 3 + \log_a 4 = \log_a 24.

It follows that a=24.a = 24.

Thus, the correct answer is D.

Problem 14 in Other Years