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

m
Line 17: Line 17:
  
 
~Technodoggo
 
~Technodoggo
 +
 +
==Solution 2 (if you're bad at logs)
 +
 +
Convert the two equations into exponents.
 +
\begin{align*}
 +
<math>x^{10}=y^x</math> \\
 +
<math>y^{10}=x^{4y}</math> \\
 +
\end{align*}
 +
  
 
==See also==
 
==See also==

Revision as of 18:41, 2 February 2024

Problem

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

==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*}


See also

2024 AIME I (ProblemsAnswer KeyResources)
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. AMC logo.png