Difference between revisions of "2015 AMC 12A Problems/Problem 14"
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==Problem== | ==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>? | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>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>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<math> \textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 36</math> | <math> \textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 36</math> | ||
Revision as of 19:14, 11 November 2015
Problem
What is the value of 
for which 
?
Solution
We use the change of base formula to show that
![\[\log_a b = \dfrac{\log_b b}{\log_b a} = \dfrac{1}{\log_b a}.\]](http://latex.artofproblemsolving.com/f/7/3/f73e271cefc08bdc7f68146a8b825b7186a45f32.png)
Thus, our equation becomes
![\[\log_a 2 + \log_a 3 + \log_a 4 = 1,\]](http://latex.artofproblemsolving.com/1/0/2/10290ea33d9b8f7e8c2fce22f914ab6872c10f8a.png)
which becomes after combining:
![\[\log_a 24 = 1.\]](http://latex.artofproblemsolving.com/c/d/3/cd3467e85a1cde569caa3051ba6a69ec140d9b84.png)
Hence 
, and the answer is 
See Also
| 2015 AMC 12A (Problems • Answer Key • Resources) | |
| 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 | |
