Difference between revisions of "2019 AMC 12A Problems/Problem 12"
Sevenoptimus (talk | contribs) (Fixed formatting and added a simpler solution) |
(Added solution 4. Simpler I think.) |
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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>. | 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>. | ||
+ | |||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Denote <math>A=\log_2 x</math> and <math>B=\log_2 y</math>. | ||
+ | |||
+ | Writing the first given as | ||
+ | <math>\log_2 x = \frac{\log_2 16}{\log_2 y}</math> and the second as | ||
+ | <math>\log_2 x + \log_2 y = \log_2 64</math>, we get <math>A\cdot B = 4</math> and <math>A+B=6</math>. | ||
+ | |||
+ | Solving for <math>B</math> we get <math>B = 3 \pm \sqrt{5}</math>. | ||
+ | |||
+ | Our goal is <math>( A-B )^2</math>. From the above it is equal to <math>(6-2B) = \left(2\sqrt{5}\right)^2 = 20 \Rightarrow \boxed{B}</math> | ||
+ | |||
==See Also== | ==See Also== |
Revision as of 19:05, 12 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
Slightly simpler solution
After obtaining , notice that the required answer is as before.
Solution 2
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 known values, we get or .
Our answer is 20 .
Solution 3
Multiplying the first equation by we obtain .
From the second equation we have .
Then, .
Solution 4
Denote and .
Writing the first given as and the second as , we get and .
Solving for we get .
Our goal is . From the above it is equal to
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 |
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