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 $\textbf{(D)}.$

Video Solution

https://youtu.be/RdIIEhsbZKw?t=482

~ pi_is_3.14

Video Solution

Video starts at 1:04, uses the above solution: https://www.youtube.com/watch?v=OJyWHzjiu2A&t=44s

See Also

2015 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 12 Problems and Solutions
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