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) | + | <math> \textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 36</math> |
==Solution== | ==Solution== | ||
| Line 12: | Line 12: | ||
which becomes after combining: | which becomes after combining: | ||
<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> | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | 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 == | ||
| + | {{AMC12 box|year=2015|ab=A|num-b=13|num-a=15}} | ||
Latest revision as of 14:46, 21 August 2024
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 
Video Solution by OmegaLearn
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 (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 | |
