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

Revision as of 19:04, 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: \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 $xy=\boxed{25}$ .

@@ Line 35: / Line 35: @@
 <cmath>{\frac{25}{x}}={x^{\frac{10}{x}}}</cmath>
-But <math>x^{\frac{10}{x}}=y</math>, so <math>\frac{25}{x}=y</math> and <math>xy=\boxed{25}</math>.
+But we know that
+<math>x^{\frac{10}{x}}=y</math>, so <math>\frac{25}{x}=y</math> and <math>xy=\boxed{25}</math>.
 ==See also==

2024 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2024 AIME I Problems/Problem 2"

Revision as of 19:04, 2 February 2024

Contents

Problem

Solution 1

Solution 2 (if you're bad at logs)

See also