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

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Plug this into (2):
 
Plug this into (2):
  
<cmath>x^{(\frac{10}{y})(10)}=x^{4x^{\frac{10}{y}}}</cmath>
+
<cmath>x^{(\frac{10}{y})(10)}=x^{4x^{\frac{10}{y}}}, so</cmath>
  
  

Revision as of 18:58, 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:

\[x^{10}=y^x~(1)\] \[y^{10}=x^{4y}~(2).\]

Take $(1)$ to the power of $\frac{1}{x}$:

\[x^{\frac{10}{y}}=y.\]

Plug this into (2):

\[x^{(\frac{10}{y})(10)}=x^{4x^{\frac{10}{y}}}, so\]


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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