Difference between revisions of "2024 AIME I Problems/Problem 2"
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− | <cmath>x\log_xy | + | <cmath>x\log_xy=10</cmath> |
− | <cmath>4y\log_yx | + | <cmath>4y\log_yx=10.</cmath> |
We multiply the two equations to get: | We multiply the two equations to get: | ||
<cmath>4xy\left(\log_xy\log_yx\right)=100.</cmath> | <cmath>4xy\left(\log_xy\log_yx\right)=100.</cmath> |
Revision as of 19:25, 11 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: $begin{align*} x\log_xy&=10 4y\log_yx&=10.$ (Error compiling LaTeX. Unknown error_msg)end{align*}
We multiply the two equations to get:
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.