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

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

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All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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