Difference between revisions of "2019 AMC 12A Problems/Problem 12"
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<math>\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32</math> | <math>\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32</math> | ||
− | ==Solution== | + | ==Solution 1 == |
− | Let <math>\log_2{x} = \log_y{16}=k</math>, | + | Let <math>\log_2{x} = \log_y{16}=k</math>, so that <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>. |
+ | We therefore have <math>k+\frac{4}{k}=6</math>, and deduce <math>k^2-6k+4=0</math>. The solutions to this are <math>k = 3 \pm \sqrt{5}</math>. | ||
− | + | To solve the problem, we now find | |
+ | <cmath>\begin{align*} | ||
+ | (\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\ | ||
+ | &= (3 \pm \sqrt{5} - 3 \pm \sqrt{5})^2\ | ||
+ | &=(\pm 2\sqrt{5})^2 \ | ||
+ | &= \boxed{\textbf{(B) } 20}. \ | ||
+ | \end{align*}</cmath> | ||
+ | ~Edits by BakedPotato66 | ||
+ | ==Solution 2 (slightly simpler)== | ||
− | + | After obtaining <math>k + \frac{4}{k} = 6</math>, notice that the required answer is <math>\left(k - \frac{4}{k}\right)^{2} = k^2 - 8 + \frac{16}{k^2} = \left(k^2 + 8 + \frac{16}{k^2}\right) - 16 = \left(k+\frac{4}{k}\right)^2 - 16 = 6^2 - 16 = \boxed{\textbf{(B) } 20}</math>, as before. | |
− | - | + | ==Solution 3== |
+ | |||
+ | From the given data, <math>\log_2(x) = \frac{1}{\log_{16}(y)}</math>, or <math>\log_2(x) = \frac{4}{{\log_{2}(y)}}</math> | ||
+ | |||
+ | We know that <math>xy=64</math>, so <math>x= \frac{64}{y}</math>. | ||
+ | |||
+ | Thus <math>\log_2\left(\frac{64}{y}\right) = \frac{4}{{\log_{2}(y)}}</math>, so <math>6-\log_2(y) = \frac{4}{{\log_{2}(y)}}</math>, so <math>6(\log_2(y))-(\log_2(y))^2=4</math>. | ||
+ | |||
+ | Solving for <math>\log_2(y)</math>, we obtain <math>\log_2(y)=3+\sqrt{5}</math>. | ||
− | == | + | Easy resubstitution further gives <math>\log_2(x)=\frac{4}{3+\sqrt{5}}</math>. Simplifying, we obtain <math>\log_2(x)= 3-\sqrt{5}</math>. |
− | + | Looking back at the original problem, we have What is <math>(\log_2{\tfrac{x}{y}})^2</math>? | |
− | + | Deconstructing this expression using log rules, we get <math>(\log_2{x}-\log_2{y})^2</math>. | |
− | + | Plugging in our known values, we get <math>((3-\sqrt{5})-(3+\sqrt{5}))^2</math> or <math>(-2\sqrt{5})^2</math>. | |
− | + | Our answer is <math>\boxed{\textbf{(B) } 20}</math>. | |
− | + | ==Solution 4== | |
− | + | Multiplying the first equation by <math>\log_2 y</math>, we obtain <math>\log_2 x\cdot\log_2 y=4</math>. | |
− | + | From the second equation we have <math>\log_2 x+\log_2 y = \log_2 (xy) = 6</math>. | |
+ | |||
+ | Then, <math>\left(\log_2 \frac{x}{y}\right)^{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 5== | |
− | + | Let <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 to find <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>. | |
− | + | Alternatively, once we found <math>AB=4</math> and <math>A+B=6</math>, we could have squared the latter to get <math>A^2+B^2+2AB=36</math>; subtracting <math>4</math> times the former equation, we find that <math>A^2+B^2-2AB=(A-B)^2=36-16=\boxed{\textbf{(B) }20}</math>. (Alternate finish by Technodoggo) | |
− | ==Solution | + | ==Video Solution 1== |
+ | https://youtu.be/ODOWgzhVKog | ||
− | + | ~Education, the Study of Everything | |
− | + | == Video Solution by OmegaLearn == | |
+ | https://youtu.be/RdIIEhsbZKw?t=1821 | ||
− | + | ~ pi_is_3.14 | |
==See Also== | ==See Also== |
Latest revision as of 00:16, 3 November 2024
Contents
[hide]Problem
Positive real numbers and satisfy and . What is ?
Solution 1
Let , so that and . Then we have .
We therefore have , and deduce . The solutions to this are .
To solve the problem, we now find ~Edits by BakedPotato66
Solution 2 (slightly simpler)
After obtaining , notice that the required answer is , as before.
Solution 3
From the given data, , or
We know that , so .
Thus , so , so .
Solving for , we obtain .
Easy resubstitution further gives . Simplifying, 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 .
Solution 4
Multiplying the first equation by , we obtain .
From the second equation we have .
Then, .
Solution 5
Let and .
Writing the first given as and the second as , we get and .
Solving for we get .
Our goal is to find . From the above, it is equal to .
Alternatively, once we found and , we could have squared the latter to get ; subtracting times the former equation, we find that . (Alternate finish by Technodoggo)
Video Solution 1
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/RdIIEhsbZKw?t=1821
~ pi_is_3.14
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.