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:
\begin{align*}
x\log_xy&=10 \\
4y\log_yx&=10. \\
\end{align*}
We multiply the two equations to get:
![\[4xy\left(\log_xy\log_yx\right)=100.\]](http://latex.artofproblemsolving.com/4/3/3/433f42f4fda44557d99bd60f5db220c4e0682a93.png)
Also by properties of logarithms, we know that 
; thus, 
. Therefore, our equation simplifies to:
![\[4xy=100\implies xy=\boxed{025}.\]](http://latex.artofproblemsolving.com/9/6/8/968ba9e7f231134a15dd1b22aca223a5d5d3947c.png)
~Technodoggo
Solution 2 (if you're bad at logs)
Convert the two equations into exponents:
![\[x^{10}=y^x\hspace{20mm} (1)\]](http://latex.artofproblemsolving.com/f/5/f/f5f412285f16bb4bdac8cd756d7a421ef90083ff.png)
![\[y^{10}=x^{4y}\hspace{20mm} (2).\]](http://latex.artofproblemsolving.com/e/2/3/e2368c4f1ee4f3782135cc1240efe75a62c4288e.png)
Take the top equation to the power of 
.
![\[x^{\frac{10}{y}}=y.\]](http://latex.artofproblemsolving.com/2/8/4/28448feb049fe14bb18677f5e754c5baf0ea105a.png)
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. 
