2022 AMC 12B Problems/Problem 16

Problem

Suppose $x$ and $y$ are positive real numbers such that

$x^y=2^{64}$ 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

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2022 AMC 12B Problems/Problem 16

Contents

Problem

Solution

Solution 2

See Also