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2019 AMC 12A Problems/Problem 12: Difference between revisions

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Plugging in our know values, we get <math>((3-\sqrt{5})-(3+\sqrt{5}))^2</math> or <math>(-2\sqrt{5})^2</math>.
Plugging in our know values, we get <math>((3-\sqrt{5})-(3+\sqrt{5}))^2</math> or <math>(-2\sqrt{5})^2</math>.


Our answer is 20 \boxed{B{}
Our answer is 20 <math>\boxed{B}</math>


==See Also==
==See Also==

Revision as of 18:33, 9 February 2019

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 $\boxed{B}$

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

These problems are copyrighted © by the Mathematical Association of America.

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