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

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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>.  
  
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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>.
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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{\textbf{(B) } 20}</math>.
  
 
==Solution 2 (slightly simpler)==
 
==Solution 2 (slightly simpler)==
  
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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>(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 = \boxed{\textbf{(B) } 20}</math>, as before.
  
 
==Solution 3==
 
==Solution 3==

Revision as of 16:39, 1 January 2021

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 $(\log_2\tfrac{x}{y})^2=(\log_2 x - \log_2 y)^2=(k-\tfrac{4}{k})^2=(3 \pm \sqrt{5} - \tfrac{4}{3 \pm \sqrt{5}})^2 = (3 \pm \sqrt{5} - 3 \mp \sqrt{5})^2= (\pm 2\sqrt{5})^2 = \boxed{\textbf{(B) } 20}$.

Solution 2 (slightly simpler)

After obtaining $k + \frac{4}{k} = 6$, notice that the required answer is $(k - \frac{4}{k})^{2} = k^2 - 8 + \frac{16}{k^2} = \left(k^2 + 8 + \frac{16}{k^2}\right) - 16 = \left(k+\frac{4}{k}\right)^2 - 16 = 6^2 - 16 = \boxed{\textbf{(B) } 20}$, 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 $\log_2(\frac{64}{y}) = \frac{4}{{\log_{2}(y)}}$, 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, $(\log_2 \frac{x}{y})^{2} = (\log_2 x-\log_2 y)^{2} = (\log_2 x+\log_2 y)^{2} - 4\log_2 x\cdot\log_2 y = (6)^{2} - 4(4) = 20 \Rightarrow \boxed{B}$.

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 $(6-2B) = \left(2\sqrt{5}\right)^2 = 20 \Rightarrow \boxed{B}$.

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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