Difference between revisions of "2024 AIME I Problems/Problem 2"

Line 16: Line 16:
  
 
~Technodoggo
 
~Technodoggo
 +
 +
==See also==
 +
{{AIME box|year=2024|n=I|num-b=1|after=3}}
 +
 +
[[Category:Intermediate Algebra Problems]]
 +
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Revision as of 12:44, 2 February 2024

There exist real numbers $x$ and $y$, both greater than 1, such that $\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10$. Find $xy$.

Solution 1

By properties of logarithms, we can simplify the given equation to $x\log_xy=4y\log_yx=10$. 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: \[4xy\left(\log_xy\log_yx\right)=100.\]

Also by properties of logarithms, we know that $\log_ab\cdot\log_ba=1$; thus, $\log_xy\cdot\log_yx=1$. Therefore, our equation simplifies to:

\[4xy=100\implies xy=\boxed{025}.\]

~Technodoggo

See also

2024 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
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. AMC logo.png