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

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 2

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:

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

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2010 AIME I Problems/Problem 3

Contents

Problem

Solution

Solution 2

See Also