2019 AMC 12A Problems/Problem 12

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

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 12 Problems and Solutions

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