2015 AMC 12A Problem 14

Problem 14 of 25IntermediateAlgebra

What is the value of

a

for which

\dfrac{1}{\log_2 a} + \dfrac{1}{\log_3 a} + \dfrac{1}{\log_4 a} = 1?

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Solution

By the change-of-base formula,

\dfrac{1}{\log_b a} = \log_a b.

Therefore

\begin{aligned} &1 = \log_a 2 + \log_a 3 \\ &\quad {}+ \log_a 4 = \log_a 24. \end{aligned}

It follows that

a = 24.

Thus, the correct answer is D.

Concepts: logarithm

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.