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

Revision as of 11:46, 24 August 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. $x^{\frac34x} = (\frac34x)^x$ $x^{\frac34x} = (\frac34)^x \cdot x^x$ $x^{-\frac14x} = (\frac34)^x$ $x^{-\frac14} = \frac34$ $x = \frac{256}{81}$ $y = \frac34x = \frac{192}{81}$ $x + y = \frac{448}{81}$ $448 + 81 = \boxed{529}$

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 (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 33: / Line 33: @@
 <center> <cmath>x+y = \left(\frac{4}{3}\right)^3 \left(\frac{4}{3} + 1 \right) = \frac{448}{81} \Longrightarrow 448 + 81 = \boxed{529}</cmath> </center>
-== Solution 4 ==
+== Solution 4 (another version of Solution 3)==
-Taking the natural logarithm of both sides, we get:
+Taking the logarithm base <math>x</math> of both sides, we arrive with:
+<cmath> y = \log_x y^x \Longrightarrow \frac{y}{x} = \log_{x} y = \log_x \left(\frac{3}{4}x\right) = \frac{3}{4}</cmath>
-<center><cmath> \ln {x^y} = \ln {y^x} </cmath></center>
+Now we proceed by the logarithm rule <math>\log(ab)=\log a + \log b</math>. The equation becomes:
+<cmath>\log_x \frac{3}{4} + \log_x x = \frac{3}{4}</cmath>
+<cmath>\Longleftrightarrow \log_x \frac{3}{4} + 1 = \frac{3}{4}</cmath>
+<cmath>\Longleftrightarrow \log_x \frac{3}{4} = -\frac{1}{4}</cmath>
+<cmath>\Longleftrightarrow x^{-\frac{1}{4}} = \frac{3}{4}</cmath>
+<cmath>\Longleftrightarrow \frac{1}{x^{\frac{1}{4}}} = \frac{3}{4}</cmath>
+<cmath>\Longleftrightarrow x^{\frac{1}{4}} = \frac{4}{3}</cmath>
+<cmath>\Longleftrightarrow \sqrt[4]{x} = \frac{4}{3}</cmath>
+<cmath>\Longleftrightarrow x = \left(\frac{4}{3}\right)^4=\frac{256}{81}</cmath>
+Then find <math>y</math> as in solution 3, and we get <math>\boxed{529}</math>.
 == See Also ==

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