Difference between revisions of "2015 AMC 12A Problems/Problem 14"
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==Problem== | ==Problem== | ||
− | <!-- 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}{\ | + | <!-- 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}{\log_2a} + \frac{1}{\log_3a} + \frac{1}{\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> | ||
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<cmath>\log_a 2 + \log_a 3 + \log_a 4 = 1,</cmath> | <cmath>\log_a 2 + \log_a 3 + \log_a 4 = 1,</cmath> | ||
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> |
Latest revision as of 15:45, 23 March 2025
Problem
What is the value of for which
?
Solution
We use the change of base formula to show that
Thus, our equation becomes
which becomes after combining:
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 |