Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 12A Problems/Problem 12"

(Fixed formatting and removed irrelevant attributions)
(→‎Solution 5)
 
(10 intermediate revisions by 6 users not shown)
Line 11: Line 11:
 
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>.  
 
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 <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 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)==
 
==Solution 2 (slightly simpler)==
  
−
After obtaining <math>k + \frac{4}{k} = 6</math>, notice that the required answer is <math>(k - \frac{4}{k})^{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 = 20</math>, as before.
+
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==
 
==Solution 3==
Line 23: Line 32:
 
We know that <math>xy=64</math>, so <math>x= \frac{64}{y}</math>.
 
We know that <math>xy=64</math>, so <math>x= \frac{64}{y}</math>.
  
−
Thus <math>\log_2(\frac{64}{y}) = \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>.
+
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>.
 
Solving for <math>\log_2(y)</math>, we obtain <math>\log_2(y)=3+\sqrt{5}</math>.
Line 35: Line 44:
 
Plugging in our known values, we get <math>((3-\sqrt{5})-(3+\sqrt{5}))^2</math> or <math>(-2\sqrt{5})^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>.
+
Our answer is <math>\boxed{\textbf{(B) } 20}</math>.
  
 
==Solution 4==
 
==Solution 4==
Line 43: Line 52:
 
From the second equation we have <math>\log_2 x+\log_2 y = \log_2 (xy) = 6</math>.
 
From the second equation we have <math>\log_2 x+\log_2 y = \log_2 (xy) = 6</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>.
+
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==
 
==Solution 5==
Line 55: Line 64:
 
Solving for <math>B</math> we get <math>B = 3 \pm \sqrt{5}</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>
+
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)
 +
 
 +
==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

Problem

Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2{x} = \log_y{16}$ and $xy = 64$. What is $(\log_2{\tfrac{x}{y}})^2$?

$\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32$

Solution 1

Let $\log_2{x} = \log_y{16}=k$, so that $2^k=x$ and $y^k=16 \implies y=2^{\frac{4}{k}}$. Then we have $(2^k)(2^{\frac{4}{k}})=2^{k+\frac{4}{k}}=2^6$.

We therefore have $k+\frac{4}{k}=6$, and deduce $k^2-6k+4=0$. The solutions to this are $k = 3 \pm \sqrt{5}$.

To solve the problem, we now find \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*} ~Edits by BakedPotato66

Solution 2 (slightly simpler)

After obtaining $k + \frac{4}{k} = 6$, notice that the required answer is $\left(k - \frac{4}{k}\right)^{2} = k^2 - 8 + \frac{16}{k^2} = \left(k^2 + 8 + \frac{16}{k^2}\right) - 16 = \left(k+\frac{4}{k}\right)^2 - 16 = 6^2 - 16 = \boxed{\textbf{(B) } 20}$, as before.

Solution 3

From the given data, $\log_2(x) = \frac{1}{\log_{16}(y)}$, or $\log_2(x) = \frac{4}{{\log_{2}(y)}}$

We know that $xy=64$, so $x= \frac{64}{y}$.

Thus $\log_2\left(\frac{64}{y}\right) = \frac{4}{{\log_{2}(y)}}$, so $6-\log_2(y) = \frac{4}{{\log_{2}(y)}}$, so $6(\log_2(y))-(\log_2(y))^2=4$.

Solving for $\log_2(y)$, we obtain $\log_2(y)=3+\sqrt{5}$.

Easy resubstitution further gives $\log_2(x)=\frac{4}{3+\sqrt{5}}$. Simplifying, we obtain $\log_2(x)= 3-\sqrt{5}$.

Looking back at the original problem, we have What is $(\log_2{\tfrac{x}{y}})^2$?

Deconstructing this expression using log rules, we get $(\log_2{x}-\log_2{y})^2$.

Plugging in our known values, we get $((3-\sqrt{5})-(3+\sqrt{5}))^2$ or $(-2\sqrt{5})^2$.

Our answer is $\boxed{\textbf{(B) } 20}$.

Solution 4

Multiplying the first equation by $\log_2 y$, we obtain $\log_2 x\cdot\log_2 y=4$.

From the second equation we have $\log_2 x+\log_2 y = \log_2 (xy) = 6$.

Then, $\left(\log_2 \frac{x}{y}\right)^{2} = (\log_2 x-\log_2 y)^{2} = (\log_2 x+\log_2 y)^{2} - 4\log_2 x\cdot\log_2 y = (6)^{2} - 4(4) = 20 \Rightarrow \boxed{B}$.

Solution 5

Let $A=\log_2 x$ and $B=\log_2 y$.

Writing the first given as $\log_2 x = \frac{\log_2 16}{\log_2 y}$ and the second as $\log_2 x + \log_2 y = \log_2 64$, we get $A\cdot B = 4$ and $A+B=6$.

Solving for $B$ we get $B = 3 \pm \sqrt{5}$.

Our goal is to find $( A-B )^2$. From the above, it is equal to $(6-2B) = \left(2\sqrt{5}\right)^2 = 20 \Rightarrow \boxed{B}$.

Alternatively, once we found $AB=4$ and $A+B=6$, we could have squared the latter to get $A^2+B^2+2AB=36$; subtracting $4$ times the former equation, we find that $A^2+B^2-2AB=(A-B)^2=36-16=\boxed{\textbf{(B) }20}$. (Alternate finish by Technodoggo)

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_12&oldid=231307"