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

Let $\log_2{x} = \log_y{16}=k$ , then $2^k=x$ 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 equate $k+\frac{4}{k}=6$ , and get $k^2-6k+4=0$ . The solutions to this are $3 \pm \sqrt{5}$ .

To solve the given, $(\log_2\tfrac{x}{y})^2=(\log_2 x - \log_2 y)^2=(k-\tfrac{4}{k})^2=(3 \pm \sqrt{5} - \tfrac{4}{3 \pm \sqrt{5}})^2 = (3 \pm \sqrt{5} - 3 \mp \sqrt{5})^2= (\pm 2\sqrt{5})^2 = \boxed{20}$

-WannabeCharmander

Thus $\log_2(x) = \frac{1}{\log_{16}(y)}.$ or $\log_2(x) = \frac{4}{{\log_{2}(y)}}$

We know that $xy=64$ .

Thus $x= \frac{64}{y}.$

Thus $\log_2(\frac{64}{y}) = \frac{4}{{\log_{2}(y)}}$

Thus $6-\log_2(y) = \frac{4}{{\log_{2}(y)}}$

Thus $6(\log_2(y))-(\log_2(y))^2=4$

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

Easy resubstitution makes $\log_2(x)=\frac{4}{3+\sqrt{5}}$

Solving for $\log_2(x)$ 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 know values, we get $((3-\sqrt{5})-(3+\sqrt{5}))^2$ or $(-2\sqrt{5})^2$ .

Our answer is 20 \boxed{B{}

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.

Page

Toolbox

Search

2019 AMC 12A Problems/Problem 12

Problem

Solution

See Also