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

Revision as of 20:52, 17 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 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 $(\log_2\tfrac{x}{y})^2=(\log_2 x - \log_2 y)^2=(k-\tfrac{4}{k})^2=(3 \pm \sqrt{5} - \tfrac{4}{3 \pm \sqrt{5}})^2 = (3 \pm \sqrt{5} - 3 \mp \sqrt{5})^2= (\pm 2\sqrt{5})^2 = \boxed{20}$ .

Solution 2 (slightly simpler)

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 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 $\log_2(\frac{64}{y}) = \frac{4}{{\log_{2}(y)}}$ , 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, $(\log_2 \frac{x}{y})^{2} = (\log_2 x-\log_2 y)^{2} = (\log_2 x+\log_2 y)^{2} - 4\log_2 x\cdot\log_2 y = (6)^{2} - 4(4) = 20 \Rightarrow \boxed{B}$ .

Solution 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 $(6-2B) = \left(2\sqrt{5}\right)^2 = 20 \Rightarrow \boxed{B}$

@@ Line 5: / Line 5: @@
 <math>\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32</math>
-==Solution==
+==Solution 1 ==
-Let <math>\log_2{x} = \log_y{16}=k</math>, then <math>2^k=x</math> and <math>y^k=16 \implies y=2^{\frac{4}{k}}</math>. Then we have <math>(2^k)(2^{\frac{4}{k}})=2^{k+\frac{4}{k}}=2^6</math>.
+Let <math>\log_2{x} = \log_y{16}=k</math>, so that <math>2^k=x</math> and <math>y^k=16 \implies y=2^{\frac{4}{k}}</math>. Then we have <math>(2^k)(2^{\frac{4}{k}})=2^{k+\frac{4}{k}}=2^6</math>.
+We therefore have <math>k+\frac{4}{k}=6</math>, and deduce <math>k^2-6k+4=0</math>. The solutions to this are <math>k = 3 \pm \sqrt{5}</math>.
-We equate <math>k+\frac{4}{k}=6</math>, and get <math>k^2-6k+4=0</math>. The solutions to this are <math>3 \pm \sqrt{5}</math>.
+To solve the problem, we now find <math>(\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}</math>.
+==Solution 2 (slightly simpler)==
-To solve the given, <math>(\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}</math>
+After obtaining <math>k + \frac{4}{k} = 6</math>, notice that the required answer is <math>(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</math>, as before.
--WannabeCharmander
+==Solution 3==
-==Slightly simpler solution==
-After obtaining <math>k + \frac{4}{k} = 6</math>, notice that the required answer is <math>(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</math> as before.
-==Solution 2==
+From the given data, <math>\log_2(x) = \frac{1}{\log_{16}(y)}</math>, or <math>\log_2(x) = \frac{4}{{\log_{2}(y)}}</math>
-Thus <math>\log_2(x) = \frac{1}{\log_{16}(y)}.</math> or <math>\log_2(x) = \frac{4}{{\log_{2}(y)}}</math>
+We know that <math>xy=64</math>, so <math>x= \frac{64}{y}</math>.
-We know that <math>xy=64</math>.
+Thus <math>\log_2(\frac{64}{y}) = \frac{4}{{\log_{2}(y)}}</math>, so <math>6-\log_2(y) = \frac{4}{{\log_{2}(y)}}</math>, so <math>6(\log_2(y))-(\log_2(y))^2=4</math>.
-Thus <math>x= \frac{64}{y}.</math>
-Thus  <math>\log_2(\frac{64}{y}) = \frac{4}{{\log_{2}(y)}}</math>
-Thus  <math>6-\log_2(y) = \frac{4}{{\log_{2}(y)}}</math>
-Thus <math>6(\log_2(y))-(\log_2(y))^2=4</math>
 Solving for <math>\log_2(y)</math>, we obtain <math>\log_2(y)=3+\sqrt{5}</math>.
-Easy resubstitution makes <math>\log_2(x)=\frac{4}{3+\sqrt{5}}</math>
+Easy resubstitution further gives <math>\log_2(x)=\frac{4}{3+\sqrt{5}}</math>. Simplifying, we obtain <math>\log_2(x)= 3-\sqrt{5}</math>.
-Solving for <math>\log_2(x)</math> we obtain <math>\log_2(x)= 3-\sqrt{5}</math>.
 Looking back at the original problem, we have What is <math>(\log_2{\tfrac{x}{y}})^2</math>?
@@ Line 47: / Line 35: @@
 Plugging in our known values, we get <math>((3-\sqrt{5})-(3+\sqrt{5}))^2</math> or <math>(-2\sqrt{5})^2</math>.
-Our answer is 20 <math>\boxed{B}</math>.
+Our answer is <math>boxed{\textbf{(B) } 20}</math>.
-==Solution 3==
+==Solution 4==
-Multiplying the first equation by <math>\log_2 y</math> we obtain <math>\log_2 x\cdot\log_2 y=4</math>.
+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>.
@@ Line 57: / Line 45: @@
 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>.
+==Solution 5==
-==Solution 4==
+Let <math>A=\log_2 x</math> and <math>B=\log_2 y</math>.
-Denote <math>A=\log_2 x</math> and <math>B=\log_2 y</math>.
 Writing the first given as
 <math>\log_2 x = \frac{\log_2 16}{\log_2 y}</math> and the second as
-<math>\log_2 x + \log_2 y = \log_2 64</math>, we get <math>A\cdot B = 4</math>   and   <math>A+B=6</math>.
+<math>\log_2 x + \log_2 y = \log_2 64</math>, we get <math>A\cdot B = 4</math> and <math>A+B=6</math>.
-Solving for <math>B</math> we get  <math>B = 3 \pm \sqrt{5}</math>.
-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>
+Solving for <math>B</math> we get <math>B = 3 \pm \sqrt{5}</math>.
---DrB_Coach.
+Our goal is to find <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>
 ==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
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