Difference between revisions of "2010 AIME I Problems/Problem 3"

Latest revision as of 21:07, 17 December 2021

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

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 (another version of Solution 3)

Taking the logarithm base $x$ of both sides, we arrive with: $y = \log_x y^x \Longrightarrow \frac{y}{x} = \log_{x} y = \log_x \left(\frac{3}{4}x\right) = \frac{3}{4}$ Now we proceed by the logarithm rule $\log(ab)=\log a + \log b$ . The equation becomes: $\log_x \frac{3}{4} + \log_x x = \frac{3}{4}$ $\Longleftrightarrow \log_x \frac{3}{4} + 1 = \frac{3}{4}$ $\Longleftrightarrow \log_x \frac{3}{4} = -\frac{1}{4}$ $\Longleftrightarrow x^{-\frac{1}{4}} = \frac{3}{4}$ $\Longleftrightarrow \frac{1}{x^{\frac{1}{4}}} = \frac{3}{4}$ $\Longleftrightarrow x^{\frac{1}{4}} = \frac{4}{3}$ $\Longleftrightarrow \sqrt[4]{x} = \frac{4}{3}$ $\Longleftrightarrow x = \left(\frac{4}{3}\right)^4=\frac{256}{81}$ Then find $y$ as in solution 3, and we get $\boxed{529}$ .

@@ Line 4: / Line 4: @@
 == Solution 1 ==
 Substitute <math>y = \frac34x</math> into <math>x^y = y^x</math> and solve.
-<cmath>x^{\frac34x} = (\frac34x)^x</cmath>
+<cmath>x^{\frac34x} = \left(\frac34x\right)^x</cmath>
-<cmath>x^{\frac34x} = (\frac34)^x \cdot x^x</cmath>
+<cmath>x^{\frac34x} = \left(\frac34\right)^x \cdot x^x</cmath>
-<cmath>x^{-\frac14x} = (\frac34)^x</cmath>
+<cmath>x^{-\frac14x} = \left(\frac34\right)^x</cmath>
 <cmath>x^{-\frac14} = \frac34</cmath>
 <cmath>x = \frac{256}{81}</cmath>

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

Page

Toolbox

Search