2015 AMC 12A Problems/Problem 14

Problem

What is the value of $a$ for which $\frac{1}{\text{log}_2a} + \frac{1}{\text{log}_3a} + \frac{1}{\text{log}_4a} = 1$ ?

$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 36$

Solution

We use the change of base formula to show that $\log_a b = \dfrac{\log_b b}{\log_b a} = \dfrac{1}{\log_b a}.$ Thus, our equation becomes $\log_a 2 + \log_a 3 + \log_a 4 = 1,$ which becomes after combining: $\log_a 24 = 1.$ Hence $a = 24$ , and the answer is $\textbf{(D)}.$

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2015 AMC 12A Problems/Problem 14

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