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 "2024 AIME I Problems/Problem 2"

(→‎Solution 1)
(→‎Solution 3)
Line 57: Line 57:
 
This means that <math>10=\frac{4xy}{10}</math>, or <math>100=4xy</math>, meaning that <math>xy=\boxed{25}</math>.
 
This means that <math>10=\frac{4xy}{10}</math>, or <math>100=4xy</math>, meaning that <math>xy=\boxed{25}</math>.
 
~MC413551
 
~MC413551
 +
 +
 +
=Solution 4==
 +
 +
The same with other solutions, we have obtained <math>x^{10}=y^x</math> and <math>y^{10}=x^{4y}</math>. Then, <math>x^{10}y^{10}=y^xx^{4y}</math>. So, <math>x^{10}=x^{4y}</math>, <math>y^{10}=y^{x}</math>. <math>x=10</math> and <math>y=2.5</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 01:29, 20 March 2024

Problem

There exist real numbers $x$ and $y$, both greater than 1, such that $\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10$. Find $xy$.

Video Solution & More by MegaMath

https://www.youtube.com/watch?v=jxY7BBe-4gU

Solution 1

By properties of logarithms, we can simplify the given equation to $x\log_xy=4y\log_yx=10$. Let us break this into two separate equations:

\[x\log_xy=10\] \[4y\log_yx=10.\] We multiply the two equations to get: \[4xy\left(\log_xy\log_yx\right)=100.\]

Also by properties of logarithms, we know that $\log_ab\cdot\log_ba=1$; thus, $\log_xy\cdot\log_yx=1$. Therefore, our equation simplifies to:

\[4xy=100\implies xy=\boxed{025}.\]

~Technodoggo

Solution 2

Convert the two equations into exponents:

\[x^{10}=y^x~(1)\] \[y^{10}=x^{4y}~(2).\]

Take $(1)$ to the power of $\frac{1}{x}$:

\[x^{\frac{10}{x}}=y.\]

Plug this into $(2)$:

\[x^{(\frac{10}{x})(10)}=x^{4(x^{\frac{10}{x}})}\] \[{\frac{100}{x}}={4x^{\frac{10}{x}}}\] \[{\frac{25}{x}}={x^{\frac{10}{x}}}=y,\]

So $xy=\boxed{025}$

~alexanderruan

Solution 3

Similar to solution 2, we have:

$x^{10}=y^x$ and $y^{10}=x^{4y}$

Take the tenth root of the first equation to get

$x=y^{\frac{x}{10}}$

Substitute into the second equation to get

$y^{10}=y^{\frac{4xy}{10}}$

This means that $10=\frac{4xy}{10}$, or $100=4xy$, meaning that $xy=\boxed{25}$. ~MC413551


Solution 4=

The same with other solutions, we have obtained $x^{10}=y^x$ and $y^{10}=x^{4y}$. Then, $x^{10}y^{10}=y^xx^{4y}$. So, $x^{10}=x^{4y}$, $y^{10}=y^{x}$. $x=10$ and $y=2.5$.

Video Solution

https://youtu.be/qLUahGcewT4

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/6C0yHp5GUBY

~Veer Mahajan

See also

2024 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2024_AIME_I_Problems/Problem_2&oldid=217181"