Difference between revisions of "2024 AIME I Problems/Problem 2"
Line 20: | Line 20: | ||
==Solution 2 (if you're bad at logs)== | ==Solution 2 (if you're bad at logs)== | ||
− | Convert the two equations into exponents | + | Convert the two equations into exponents: |
\begin{align*} | \begin{align*} | ||
x^{10}=y^x\\ | x^{10}=y^x\\ | ||
− | y^{10}=x^{4y}\\ | + | y^{10}=x^{4y}.\\ |
+ | \end{align*} | ||
+ | Take the top equation to the power of <math>\frac{1}{x}</math>. | ||
+ | \begin{align*} | ||
+ | x^{\frac{10}{y}}=y. | ||
\end{align*} | \end{align*} | ||
− | |||
==See also== | ==See also== |
Revision as of 18:47, 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: \begin{align*} x\log_xy&=10 \\ 4y\log_yx&=10. \\ \end{align*} We multiply the two equations to get:
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: \begin{align*} x^{10}=y^x\\ y^{10}=x^{4y}.\\ \end{align*} Take the top equation to the power of . \begin{align*} x^{\frac{10}{y}}=y. \end{align*}
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.