Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 12B Problems/Problem 16"

(Created page with "== Problem == Suppose <math>x</math> and <math>y</math> are positive real numbers such that <math>x^y=2^{64}</math> and <math>(\log_2{x})^{\log_2{y}}=2^{27}</math>. What is...")
 
(solution added)
Line 20: Line 20:
  
 
cr. djmathman
 
cr. djmathman
 +
 +
== Solution 2 ==
 +
 +
<math>x^y=2^{64} \Rightarrow y\log_2{x}=64 \Rightarrow \log_2{x}=\dfrac{64}{y}</math>.
 +
 +
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>.
 +
 +
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>.
 +
 +
~4SunnyH
  
 
== 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}}

Revision as of 00:21, 18 November 2022

Problem

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

$x^y=2^{64}$ and $(\log_2{x})^{\log_2{y}}=2^{27}$.

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12B_Problems/Problem_16&oldid=182169"