2024 AIME I Problems/Problem 2

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:

$x^{10}=y^x~(1)$ $y^{10}=x^{4y}~(2).$

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

$x^{\frac{10}{x}}=y.$

Plug this into $(2)$ :

$x^{(\frac{10}{x})(10)}=x^{4x^{\frac{10}{x}}}$ ${\frac{100}{x}}={4(x^{\frac{10}{x}})}$ ${\frac{25}{x}}={x^{\frac{10}{x}}}$

But we know that $x^{\frac{10}{x}}=y$ , so $\frac{25}{x}=y$ and $xy=\boxed{25}$ .

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

Page

Toolbox

Search

2024 AIME I Problems/Problem 2

Contents

Problem

Solution 1

Solution 2 (if you're bad at logs)

See also