Difference between revisions of "2024 AIME I Problems/Problem 2"
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==Problem== | ==Problem== | ||
− | There exist real numbers <math>x</math> and <math>y</math>, both greater than 1, such that <math>\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10</math>. Find <math>xy</math>. | + | There exist real numbers <math>x</math> and <math>y</math>, both greater than 1, such that <math>\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10</math>. Find <math>xy</math>. |
+ | |||
+ | ==Video Solution & More by MegaMath== | ||
+ | https://www.youtube.com/watch?v=jxY7BBe-4gU | ||
==Solution 1== | ==Solution 1== |
Revision as of 12:54, 9 February 2024
Contents
[hide]Problem
There exist real numbers and , both greater than 1, such that . Find .
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 . Let us break this into two separate equations:
Also by properties of logarithms, we know that ; thus, . Therefore, our equation simplifies to:
~Technodoggo
Solution 2
Convert the two equations into exponents:
Take to the power of :
Plug this into :
So
~alexanderruan
Solution 3
Similar to solution 2, we have:
and
Take the tenth root of the first equation to get
Substitute into the second equation to get
This means that , or , meaning that . ~MC413551
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~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.