Difference between revisions of "2024 AIME I Problems/Problem 2"
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==Solution 6== | ==Solution 6== | ||
− | Put <cmath> y=x^a </cmath> | + | Put <cmath> y=x^a </cmath>.We see: <cmath>ax=10 </cmath> and <cmath>4x^a/a=10 </cmath> |
− | We see: <cmath>ax=10 </cmath> and <cmath>4x^a/a=10 </cmath> | + | which gives rise to <cmath>xy=25 </cmath> which is the required answer. |
− | which gives rise to <cmath> | ||
~Grammaticus | ~Grammaticus | ||
− | |||
==Video Solution== | ==Video Solution== |
Revision as of 11:50, 5 June 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
Solution 6
Put .We see: and which gives rise to which is the required answer.
~Grammaticus
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.