Difference between revisions of "2024 AIME I Problems/Problem 2"
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~alexanderruan | ~alexanderruan | ||
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+ | ==Solution 3== | ||
+ | |||
+ | Similar to solution 2, we have: | ||
+ | |||
+ | <math>x^{10}=y^x</math> and <math>y^{10}=x^{4y}</math> | ||
+ | |||
+ | Take the tenth root of the first equation to get | ||
+ | |||
+ | <math>x=y^{\frac{x}{10}}</math> | ||
+ | |||
+ | Substitute into the second equation to get | ||
+ | |||
+ | <math>y^{10}=y^{\frac{4xy}{10}}</math> | ||
+ | |||
+ | This means that <math>10=\frac{4xy}{10}</math>, or <math>100=4xy</math>, meaning that <math>xy=\boxed{25}</math>. | ||
+ | ~MC413551 | ||
==Video Solution== | ==Video Solution== |
Revision as of 15:13, 3 February 2024
Problem
There exist real numbers and , both greater than 1, such that . Find .
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)
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.