AIME 2024 I · 第 2 题

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:

$$x\log_xy=10$$ $$4y\log_yx=10.$$ 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

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 $(2)$:

$$x^{(\frac{10}{x})(10)}=x^{4(x^{\frac{10}{x}})}$$ $${\frac{100}{x}}={4x^{\frac{10}{x}}}$$ $${\frac{25}{x}}={x^{\frac{10}{x}}}=y,$$ So $xy=\boxed{025}$

~alexanderruan

Solution 3

Similar to solution 2, we have:

$x^{10}=y^x$ and $y^{10}=x^{4y}$

Take the tenth root of the first equation to get

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

Substitute into the second equation to get

$y^{10}=y^{\frac{4xy}{10}}$

This means that $10=\frac{4xy}{10}$, or $100=4xy$, meaning that $xy=\boxed{025}$.

~MC413551

Solution 4

The same with other solutions, we have obtained $x^{10}=y^x$ and $y^{10}=x^{4y}$. Then, $x^{10}y^{10}=y^xx^{4y}$. So, an obvious solution is to have $x^{10}=x^{4y}$ and $y^{10}=y^{x}$. Solving, we get $x=10$ and $y=2.5$.So $xy = \boxed{025}$.

Note: This is not a correct solution. Plugging in $x=10$ and $y=2.5$ does not satisfy the equations.

Solution 5(fakesolve)

Using the first expression, we see that $x^{10} = y^x$. Now, taking the log of both sides, we get $\log_y(x^{10}) = \log_y(y^x)$. This simplifies to $10 \log_y(x) = x$. This is still equal to the second equation in the problem statement, so $10 \log_y(x) = x = 4y \log_y(x)$. Dividing by $\log_y(x)$ on both sides, we get $x = 4y = 10$. Therefore, $x = 10$ and $y = 2.5$, so $xy = \boxed{025}$.

~idk12345678

Solution 6

Let

$$y=x^a$$ .We see:

$$ax=10$$ and

$$4x^a/a=10$$ which gives rise to

$$xy = \boxed{025}$$ .

~Grammaticus

Video Solution

https://youtu.be/qLUahGcewT4

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

Video Solution

https://youtu.be/6C0yHp5GUBY

~Veer Mahajan

Video Solution

https://youtu.be/5wHEa9Qwe3k ~Ajeet Dubey (https://www.ioqm.in)

Video Solution & More by MegaMath

https://www.youtube.com/watch?v=jxY7BBe-4gU

Video Solution By MathTutorZhengFrSG

https://youtu.be/HbGlIki_BsY

~MathTutorZhengFrSG