# 2022 AMC 12B Problems/Problem 16

## 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

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 $3+\sqrt 2 \implies \boxed{\textbf {(C)}}$.

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

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

~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

~r00tsOfUnity

## Video Solution (Just 2 min!)

Education, the Study of Everything

## Video Solution(1-16)

~~Hayabusa1

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