# 2010 AIME I Problems/Problem 3

## Problem

Suppose that $y = \frac34x$ and $x^y = y^x$. The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$, where $r$ and $s$ are relatively prime positive integers. Find $r + s$.

## Solution 1

Substitute $y = \frac34x$ into $x^y = y^x$ and solve. $$x^{\frac34x} = (\frac34x)^x$$ $$x^{\frac34x} = (\frac34)^x \cdot x^x$$ $$x^{-\frac14x} = (\frac34)^x$$ $$x^{-\frac14} = \frac34$$ $$x = \frac{256}{81}$$ $$y = \frac34x = \frac{192}{81}$$ $$x + y = \frac{448}{81}$$ $$448 + 81 = \boxed{529}$$

## Solution 2

We solve in general using $c$ instead of $3/4$. Substituting $y = cx$, we have: $$x^{cx} = (cx)^x \Longrightarrow (x^x)^c = c^x\cdot x^x$$

Dividing by $x^x$, we get $(x^x)^{c - 1} = c^x$.

Taking the $x$th root, $x^{c - 1} = c$, or $x = c^{1/(c - 1)}$.

In the case $c = \frac34$, $x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81}$, $y = \frac {64}{27}$, $x + y = \frac {256 + 192}{81} = \frac {448}{81}$, yielding an answer of $448 + 81 = \boxed{529}$.

## Solution 3

Taking the logarithm base $x$ of both sides, we arrive with: $$y = \log_x y^x \Longrightarrow \frac{y}{x} = \log_{x} y = \log_x \frac{3}{4}x = \frac{3}{4}$$

Where the last two simplifications were made since $y = \frac{3}{4}x$. Then, $$x^{\frac{3}{4}} = \frac{3}{4}x \Longrightarrow x^{\frac{1}{4}} = \frac{4}{3} \Longrightarrow x = \left(\frac{4}{3}\right)^4$$

Then, $y = \left(\frac{4}{3}\right)^3$, and thus: $$x+y = \left(\frac{4}{3}\right)^3 \left(\frac{4}{3} + 1 \right) = \frac{448}{81} \Longrightarrow 448 + 81 = \boxed{529}$$

## Solution 4

Taking the natural logarithm of both sides, we get: $$\ln {x^y} = \ln {y^x}$$

## See Also

 2010 AIME I (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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