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

Revision as of 19:23, 12 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

Slightly simpler solution

After obtaining $k + \frac{4}{k} = 6$ , notice that the required answer is $(k - \frac{4}{k})^{2} = k^2 - 8 + \frac{16}{k^2} = \left(k^2 + 8 + \frac{16}{k^2}\right) - 16 = \left(k+\frac{4}{k}\right)^2 - 16 = 6^2 - 16 = 20$ as before.

Solution 2

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

Solution 4

Denote $A=\log_2 x$ 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 $( A-B )^2$ . From the above it is equal to $(6-2B) = \left(2\sqrt{5}\right)^2 = 20 \Rightarrow \boxed{B}$

++DrB_Coach.

@@ Line 70: / Line 70: @@
 Our goal is <math>( A-B )^2</math>. From the above it is equal to <math>(6-2B) = \left(2\sqrt{5}\right)^2 = 20 \Rightarrow \boxed{B}</math>
-DrB_Coach.
+++DrB_Coach.
 ==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

Page

Toolbox

Search