2015 AMC 12A Problem 14

Attempt Problem 14 of the 2015 AMC 12A 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 2015 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

Answer: D
Concepts:logarithm
Difficulty rating: 1730
Solution:

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

It follows that a=24.a = 24.

Thus, the correct answer is D.

← Problem 13#13
Full Exam

Problem 14 in Other Years