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

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==Problem 14==
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==Problem==
  
 
What is the value of <math>a</math> for which <math>\frac{1}{\text{log}_2a} + \frac{1}{\text{log}_3a} + \frac{1}{\text{log}_4a} = 1</math>?
 
What is the value of <math>a</math> for which <math>\frac{1}{\text{log}_2a} + \frac{1}{\text{log}_3a} + \frac{1}{\text{log}_4a} = 1</math>?
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<cmath>\log_a 24 = 1.</cmath>
 
<cmath>\log_a 24 = 1.</cmath>
 
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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== See Also ==
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{{AMC12 box|year=2015|ab=A|num-b=13|num-a=15}}

Revision as of 00:50, 5 February 2015

Problem

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

$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}}\ 24\qquad\textbf{(E)}\ 36$ (Error compiling LaTeX. Unknown error_msg)

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 $\textbf{(D)}.$

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