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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>
  
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==Solution==
+
==Solution 1 ==
  
−
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>, 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>.
  
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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>.
+
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)==
  
−
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>
+
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.
  
−
-WannabeCharmander
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==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>?
  
−
Thus <math>\log_2(x) = \frac{1}{\log_{16}(y)}.</math> or <math>\log_2(x) = \frac{4}{{\log_{2}(y)}}</math>
+
Deconstructing this expression using log rules, we get <math>(\log_2{x}-\log_2{y})^2</math>.
  
−
We know that <math>xy=64</math>.
+
Plugging in our known values, we get <math>((3-\sqrt{5})-(3+\sqrt{5}))^2</math> or <math>(-2\sqrt{5})^2</math>.
  
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Thus <math>x= \frac{64}{y}.</math>
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Our answer is <math>\boxed{\textbf{(B) } 20}</math>.
  
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Thus  <math>\log_2(\frac{64}{y}) = \frac{4}{{\log_{2}(y)}}</math>
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==Solution 4==
  
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Thus  <math>6-\log_2(y) = \frac{4}{{\log_{2}(y)}}</math>
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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>.
  
−
Thus <math>6(\log_2(y))-(\log_2(y))^2=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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Solving for <math>\log_2(y)</math>, we obtain <math>\log_2(y)=3+\sqrt{5}</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>.
  
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Easy resubstitution makes <math>\log_2(x)=\frac{4}{3+\sqrt{5}}</math>
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==Solution 5==
  
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Solving for <math>\log_2(x)</math> we obtain <math>\log_2(x)= 3-\sqrt{5}</math>.
+
Let <math>A=\log_2 x</math> and <math>B=\log_2 y</math>.  
  
−
Looking back at the original problem, we have What is <math>(\log_2{\tfrac{x}{y}})^2</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>.
  
−
Deconstructing this expression using log rules, we get <math>(\log_2{x}-\log_2{y})^2</math>.
+
Solving for <math>B</math> we get <math>B = 3 \pm \sqrt{5}</math>.
  
−
Plugging in our know values, we get <math>((3-\sqrt{5})-(3+\sqrt{5}))^2</math> or <math>(-2\sqrt{5})^2</math>.
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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>.
  
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Our answer is 20 <math>\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)
  
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==Solution 3==
+
==Video Solution 1==
 +
https://youtu.be/ODOWgzhVKog
  
−
Multiplying the first equation by <math>\log_2 y</math> we obtain <math>\log_2 x\cdot\log_2 y=4</math>.
+
~Education, the Study of Everything
  
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From the second equation we have <math>\log_2 x+\log_2 y = \log_2 (xy) = 6</math>.
+
== Video Solution by OmegaLearn ==
 +
https://youtu.be/RdIIEhsbZKw?t=1821
  
−
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>.
+
~ 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

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