Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 42: Line 42:
  
 
Our answer is 20 <math>\boxed{B}</math>
 
Our answer is 20 <math>\boxed{B}</math>
 +
 +
==Solution 3==
 +
 +
Multiplying the first equation by <math>\log_2 y</math> we obtain <math>\log_2 x\cdot\log_2 y=4</math>.
 +
 +
From the second equation we have <math>\log_2 x+\log_2 y = \log_2 (xy) = 6</math>.
 +
 +
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>.
  
 
==See Also==
 
==See Also==

Revision as of 19:48, 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 $\boxed{B}$

Solution 3

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, $(\log_2 \frac{x}{y})^{2} = (\log_2 x-\log_2 y)^{2} = (\log_2 x+\log_2 y)^{2} - 4\log_2 x\cdot\log_2 y = (6)^{2} - 4(4) = 20 \Rightarrow \boxed{B}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_12&oldid=101663"