Difference between revisions of "2024 AIME I Problems/Problem 2"
Line 22: | Line 22: | ||
Convert the two equations into exponents: | Convert the two equations into exponents: | ||
− | <cmath>x^{10}=y^x</cmath> | + | <cmath>x^{10}=y^x\hspace{20mm} (1)</cmath> |
− | <cmath>y^{10}=x^{4y}.</cmath> | + | <cmath>y^{10}=x^{4y}\hspace{20mm} (2).</cmath> |
Take the top equation to the power of <math>\frac{1}{x}</math>. | Take the top equation to the power of <math>\frac{1}{x}</math>. | ||
− | <cmath>x^{\frac{10}{y}}=y</cmath> | + | <cmath>x^{\frac{10}{y}}=y.</cmath> |
− | + | Plug this into | |
==See also== | ==See also== |
Revision as of 18:50, 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 the top equation to the power of .
Plug this into
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.