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

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 $(2)$ :

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

So $xy=\boxed{025}$

~alexanderruan

Solution 3

Video Solution

https://youtu.be/qLUahGcewT4

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/6C0yHp5GUBY

~Veer Mahajan

2024 AIME I (Problems • Answer Key • Resources)
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2024 AIME I Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

Solution 3

Video Solution

Video Solution

See also