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2022 AMC 12B Problems/Problem 16: Difference between revisions

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Suppose <math>x</math> and <math>y</math> are positive real numbers such that
Suppose <math>x</math> and <math>y</math> are positive real numbers such that
 
<cmath>x^y=2^{64}\text{ and }(\log_2{x})^{\log_2{y}}=2^{7}.</cmath>
<math>x^y=2^{64}</math> and <math>(\log_2{x})^{\log_2{y}}=2^{27}</math>.
 
What is the greatest possible value of <math>\log_2{y}</math>?
What is the greatest possible value of <math>\log_2{y}</math>?


<math>\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }3+\sqrt{2} \qquad \textbf{(D) }4+\sqrt{3} \qquad \textbf{(E) }7</math>
<math>\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }3+\sqrt{2} \qquad \textbf{(D) }4+\sqrt{3} \qquad \textbf{(E) }7</math>


== Solution ==
== Solution 1 ==


Take the base-two logarithm of both equations to get
Take the base-two logarithm of both equations to get
Line 17: Line 15:
It follows that the real numbers <math>r:=\log_2 y</math> and <math>s:=\log_2\log_2 x</math> satisfy <math>r+s=6</math> and <math>rs = 7</math>. Solving this system yields
It follows that the real numbers <math>r:=\log_2 y</math> and <math>s:=\log_2\log_2 x</math> satisfy <math>r+s=6</math> and <math>rs = 7</math>. Solving this system yields
<cmath>\{\log_2 y,\log_2\log_2 x\}\in\{3-\sqrt 2, 3 + \sqrt 2\}.</cmath>
<cmath>\{\log_2 y,\log_2\log_2 x\}\in\{3-\sqrt 2, 3 + \sqrt 2\}.</cmath>
Thus the largest possible value of <math>\log_2 y</math> is <math>3+\sqrt 2 \implies \boxed{\textbf {(C)}}</math>.
Thus the largest possible value of <math>\log_2 y</math> is <math>3+\sqrt 2 \implies \boxed{\textbf{(C) }3+\sqrt{2}}</math>.


cr. djmathman
~cr. djmathman


== Solution 2 ==
== Solution 2 ==
Line 27: Line 25:
Substitution into <math>(\log_2{x})^{\log_2{y}}=2^{7}</math> yields
Substitution into <math>(\log_2{x})^{\log_2{y}}=2^{7}</math> yields


<math>(\dfrac{64}{y})^{\log_2{y}}=2^{7} \Rightarrow \log_2{y}\log_2{\dfrac{64}{y}}=7 \Rightarrow \log_2{y}(6-\log_2{y})=7 \Rightarrow \log^2_2{y}-6\log_2{y}+7=0</math>.
<math>\left(\dfrac{64}{y}\right)^{\log_2{y}}=2^{7} \Rightarrow \log_2{y}\log_2{\dfrac{64}{y}}=7 \Rightarrow \log_2{y}(6-\log_2{y})=7 \Rightarrow \log^2_2{y}-6\log_2{y}+7=0</math>.


Solving for <math>\log_2{y}</math> yields <math>\log_2{y}=3-\sqrt{2}</math> or <math>3+\sqrt{2}</math>, and we take the greater value <math>\boxed{\boldsymbol{(\textbf{C})3+\sqrt{2}}}</math>.
Solving for <math>\log_2{y}</math> yields <math>\log_2{y}=3-\sqrt{2}</math> or <math>3+\sqrt{2}</math>, and we take the greater value <math>\boxed{\textbf{(C) }3+\sqrt{2}}</math>.


~4SunnyH
~4SunnyH
== Solution 3 ==
Let <math>x = 2^a, y = 2^b.</math> We have <math>(2^a)^{2^b} = 2^{64} \Rightarrow 2^{a\cdot 2^b} = 2^{64} \Rightarrow a\cdot 2^b = 64,</math> and <math>a^b = 128</math>.
Then, from eq 1, <math>a = 64\cdot 2^{-b},</math> and substituting in to eq 2, <math>(64\cdot 2^{-b})^b = 64^b\cdot 2^{-b^2} = 2^{6b}\cdot 2^{-b^2} = 2^{6b-b^2} = 2^{7}.</math> Thus, <math>6b-b^2 = 7.</math>
Solving for <math>b</math> using the quadratic formula gets <math>b = 3 \pm \sqrt{2}.</math> Since we are looking for <math>\log_2{y}</math> which equals <math>b,</math> we put <math>\boxed{\textbf{(C) }3+\sqrt{2}}</math> as our answer.
~sirswagger21
==Video Solution by mop 2024==
https://youtu.be/ezGvZgBLe8k&t=722s
~r00tsOfUnity
==Video Solution (Just 2 min!)==
https://youtu.be/WU15Q_b9EDI
<i>Education, the Study of Everything</i>
==Video Solution(1-16)==
https://youtu.be/SCwQ9jUfr0g
~~Hayabusa1


== See Also ==
== See Also ==
{{AMC12 box|year=2022|ab=B|num-b=15|num-a=17}}
{{AMC12 box|year=2022|ab=B|num-b=15|num-a=17}}
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 03:02, 4 November 2025

Problem

Suppose $x$ and $y$ are positive real numbers such that \[x^y=2^{64}\text{ and }(\log_2{x})^{\log_2{y}}=2^{7}.\] What is the greatest possible value of $\log_2{y}$?

$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }3+\sqrt{2} \qquad \textbf{(D) }4+\sqrt{3} \qquad \textbf{(E) }7$

Solution 1

Take the base-two logarithm of both equations to get \[y\log_2 x = 64\quad\text{and}\quad (\log_2 y)(\log_2\log_2 x) = 7.\] Now taking the base-two logarithm of the first equation again yields \[\log_2 y + \log_2\log_2 x = 6.\] It follows that the real numbers $r:=\log_2 y$ and $s:=\log_2\log_2 x$ satisfy $r+s=6$ and $rs = 7$. Solving this system yields \[\{\log_2 y,\log_2\log_2 x\}\in\{3-\sqrt 2, 3 + \sqrt 2\}.\] Thus the largest possible value of $\log_2 y$ is $3+\sqrt 2 \implies \boxed{\textbf{(C) }3+\sqrt{2}}$.

~cr. djmathman

Solution 2

$x^y=2^{64} \Rightarrow y\log_2{x}=64 \Rightarrow \log_2{x}=\dfrac{64}{y}$.

Substitution into $(\log_2{x})^{\log_2{y}}=2^{7}$ yields

$\left(\dfrac{64}{y}\right)^{\log_2{y}}=2^{7} \Rightarrow \log_2{y}\log_2{\dfrac{64}{y}}=7 \Rightarrow \log_2{y}(6-\log_2{y})=7 \Rightarrow \log^2_2{y}-6\log_2{y}+7=0$.

Solving for $\log_2{y}$ yields $\log_2{y}=3-\sqrt{2}$ or $3+\sqrt{2}$, and we take the greater value $\boxed{\textbf{(C) }3+\sqrt{2}}$.

~4SunnyH

Solution 3

Let $x = 2^a, y = 2^b.$ We have $(2^a)^{2^b} = 2^{64} \Rightarrow 2^{a\cdot 2^b} = 2^{64} \Rightarrow a\cdot 2^b = 64,$ and $a^b = 128$.

Then, from eq 1, $a = 64\cdot 2^{-b},$ and substituting in to eq 2, $(64\cdot 2^{-b})^b = 64^b\cdot 2^{-b^2} = 2^{6b}\cdot 2^{-b^2} = 2^{6b-b^2} = 2^{7}.$ Thus, $6b-b^2 = 7.$

Solving for $b$ using the quadratic formula gets $b = 3 \pm \sqrt{2}.$ Since we are looking for $\log_2{y}$ which equals $b,$ we put $\boxed{\textbf{(C) }3+\sqrt{2}}$ as our answer.

~sirswagger21

Video Solution by mop 2024

https://youtu.be/ezGvZgBLe8k&t=722s

~r00tsOfUnity

Video Solution (Just 2 min!)

https://youtu.be/WU15Q_b9EDI

Education, the Study of Everything

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

See Also

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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

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