Difference between revisions of "2024 AIME I Problems/Problem 2"
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<cmath>x^{(\frac{10}{x})(10)}=x^{4x^{\frac{10}{x}}}</cmath> | <cmath>x^{(\frac{10}{x})(10)}=x^{4x^{\frac{10}{x}}}</cmath> | ||
<cmath>{\frac{100}{x}}={4(x^{\frac{10}{x}})}</cmath> | <cmath>{\frac{100}{x}}={4(x^{\frac{10}{x}})}</cmath> | ||
− | <cmath>{\frac{25}{x}}={x^{\frac{10}{x}}}</cmath> | + | <cmath>{\frac{25}{x}}={x^{\frac{10}{x}}}=y,</cmath> |
− | + | So <math>xy=\boxed{025}</math> | |
− | + | ~alexanderruan | |
==See also== | ==See also== |
Revision as of 19:06, 2 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 (if you're bad at logs)
Convert the two equations into exponents:
Take to the power of :
Plug this into :
So ~alexanderruan
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.