Difference between revisions of "2015 AMC 12A Problems/Problem 14"

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Hence <math>a = 24</math>, and the answer is <math>\textbf{(D)}.</math>
 
Hence <math>a = 24</math>, and the answer is <math>\textbf{(D)}.</math>
  
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==Video Solution==
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Video starts at 1:04, uses the above solution: https://www.youtube.com/watch?v=OJyWHzjiu2A&t=44s
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2015|ab=A|num-b=13|num-a=15}}
 
{{AMC12 box|year=2015|ab=A|num-b=13|num-a=15}}

Revision as of 17:41, 7 August 2020

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

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
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