Difference between revisions of "2019 AMC 12A Problems/Problem 12"
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Let <math>\log_2{x} = \log_y{16}=k</math>, then <math>2^k=x</math> and <math>y^k=16 \implies y=2^{\frac{4}{k}}</math>. Then we have <math>(2^k)(2^{\frac{4}{k}})=2^{k+\frac{4}{k}}=2^6</math>. | Let <math>\log_2{x} = \log_y{16}=k</math>, then <math>2^k=x</math> and <math>y^k=16 \implies y=2^{\frac{4}{k}}</math>. Then we have <math>(2^k)(2^{\frac{4}{k}})=2^{k+\frac{4}{k}}=2^6</math>. | ||
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We equate <math>k+\frac{4}{k}=6</math>, and get <math>k^2-6k+4=0</math>. The solutions to this are <math>3 \pm \sqrt{5}</math>. | We equate <math>k+\frac{4}{k}=6</math>, and get <math>k^2-6k+4=0</math>. The solutions to this are <math>3 \pm \sqrt{5}</math>. | ||
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To solve the given, <math>(\log_2\tfrac{x}{y})^2=(\log_2 x - \log_2 y)^2=(k-\tfrac{4}{k})^2=(3 \pm \sqrt{5} - \tfrac{4}{3 \pm \sqrt{5}})^2 = (3 \pm \sqrt{5} - 3 \mp \sqrt{5})^2= (\pm 2\sqrt{5})^2 = \boxed{20}</math> | To solve the given, <math>(\log_2\tfrac{x}{y})^2=(\log_2 x - \log_2 y)^2=(k-\tfrac{4}{k})^2=(3 \pm \sqrt{5} - \tfrac{4}{3 \pm \sqrt{5}})^2 = (3 \pm \sqrt{5} - 3 \mp \sqrt{5})^2= (\pm 2\sqrt{5})^2 = \boxed{20}</math> | ||
Revision as of 18:14, 9 February 2019
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
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. 
