Difference between revisions of "2019 AMC 12A Problems/Problem 12"
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Our answer is 20 <math>\boxed{B}</math> | Our answer is 20 <math>\boxed{B}</math> | ||
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+ | ==Solution 3== | ||
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+ | Multiplying the first equation by <math>\log_2 y</math> we obtain <math>\log_2 x\cdot\log_2 y=4</math>. | ||
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+ | From the second equation we have <math>\log_2 x+\log_2 y = \log_2 (xy) = 6</math>. | ||
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+ | Then, <math>(\log_2 \frac{x}{y})^{2} = (\log_2 x-\log_2 y)^{2} = (\log_2 x+\log_2 y)^{2} - 4\log_2 x\cdot\log_2 y = (6)^{2} - 4(4) = 20 \Rightarrow \boxed{B}</math>. | ||
==See Also== | ==See Also== |
Revision as of 19:48, 9 February 2019
Contents
[hide]Problem
Positive real numbers and satisfy and . What is ?
Solution
Let , then and . Then we have .
We equate , and get . The solutions to this are .
To solve the given,
-WannabeCharmander
Thus or
We know that .
Thus
Thus
Thus
Thus
Solving for , we obtain .
Easy resubstitution makes
Solving for we obtain .
Looking back at the original problem, we have What is ?
Deconstructing this expression using log rules, we get .
Plugging in our know values, we get or .
Our answer is 20
Solution 3
Multiplying the first equation by we obtain .
From the second equation we have .
Then, .
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.