Difference between revisions of "2024 AIME I Problems/Problem 2"
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==Solution 5== | ==Solution 5== | ||
− | Using the first expression, we see that <math>x^{10} = y^x</math>. Now, taking the log of both sides, we get <math>\log_y(x^{10}) = \log_y(y^x)</math>. This simplifies to <math>10 \log_y(x) = x</math>. This is still equal to the second equation in the problem statement, so <math>10 \log_y(x) = x = 4y \log_y(x)</math>. Dividing by <math>\log_y(x)</math> on both sides, we get <math>x = 4y = 10</math>. Therefore, <math>x = 10</math> and <math>y = 2.5</math>, so <math>xy = \boxed{25}</math>. | + | Using the first expression, we see that <math>x^{10} = y^x</math>. Now, taking the log of both sides, we get <math>\log_y(x^{10}) = \log_y(y^x)</math>. This simplifies to <math>10 \log_y(x) = x</math>. This is still equal to the second equation in the problem statement, so <math>10 \log_y(x) = x = 4y \log_y(x)</math>. Dividing by <math>\log_y(x)</math> on both sides, we get <math>x = 4y = 10</math>. Therefore, <math>x = 10</math> and <math>y = 2.5</math>, so <math>xy = \boxed{25}</math>. |
+ | ~idk12345678 | ||
==Video Solution== | ==Video Solution== |
Revision as of 13:02, 2 April 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:
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
Solution 4
The same with other solutions, we have obtained and . Then, . So, , . and .
Solution 5
Using the first expression, we see that . Now, taking the log of both sides, we get . This simplifies to . This is still equal to the second equation in the problem statement, so . Dividing by on both sides, we get . Therefore, and , so .
~idk12345678
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.