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 $x$ 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: xlogxy=104ylogyx=10. 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: x10=yxy10=x4y. Take the top equation to the power of $\frac{1}{x}$. x10y=y.

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

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