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

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== Problem ==
 
== Problem ==
 +
Suppose that <math>y = \frac34x</math> and <math>x^y = y^x</math>. The quantity <math>x + y</math> can be expressed as a rational number <math>\frac {r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r + s</math>.
  
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== See also ==
+
== Solution 1 ==
 +
Substitute <math>y = \frac34x</math> into <math>x^y = y^x</math> and solve.
 +
<cmath>x^{\frac34x} = \left(\frac34x\right)^x</cmath>
 +
<cmath>x^{\frac34x} = \left(\frac34\right)^x \cdot x^x</cmath>
 +
<cmath>x^{-\frac14x} = \left(\frac34\right)^x</cmath>
 +
<cmath>x^{-\frac14} = \frac34</cmath>
 +
<cmath>x = \frac{256}{81}</cmath>
 +
<cmath>y = \frac34x = \frac{192}{81}</cmath>
 +
<cmath>x + y = \frac{448}{81}</cmath>
 +
<cmath>448 + 81 = \boxed{529}</cmath>
 +
 
 +
== Solution 2 ==
 +
We solve in general using <math>c</math> instead of <math>3/4</math>. Substituting <math>y = cx</math>, we have:
 +
 
 +
<center><cmath>x^{cx} = (cx)^x \Longrightarrow (x^x)^c = c^x\cdot x^x</cmath></center>
 +
 
 +
Dividing by <math>x^x</math>, we get <math>(x^x)^{c - 1} = c^x</math>.
 +
 
 +
Taking the <math>x</math>th root, <math>x^{c - 1} = c</math>, or <math>x = c^{1/(c - 1)}</math>.
 +
 
 +
In the case <math>c = \frac34</math>, <math>x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81}</math>, <math>y = \frac {64}{27}</math>, <math>x + y = \frac {256 + 192}{81} = \frac {448}{81}</math>, yielding an answer of <math>448 + 81 = \boxed{529}</math>.
 +
 
 +
== Solution 3 ==
 +
Taking the logarithm base <math>x</math> of both sides, we arrive with:
 +
 
 +
<center><cmath> y = \log_x y^x \Longrightarrow \frac{y}{x} = \log_{x} y = \log_x \frac{3}{4}x = \frac{3}{4}</cmath></center>
 +
Where the last two simplifications were made since <math>y = \frac{3}{4}x</math>. Then,
 +
<center><cmath>x^{\frac{3}{4}} = \frac{3}{4}x \Longrightarrow x^{\frac{1}{4}} = \frac{4}{3} \Longrightarrow x = \left(\frac{4}{3}\right)^4</cmath></center>
 +
Then, <math>y = \left(\frac{4}{3}\right)^3</math>, and thus:
 +
<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 (another version of Solution 3)==
 +
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>
 +
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 ==
 
{{AIME box|year=2010|num-b=2|num-a=4|n=I}}
 
{{AIME box|year=2010|num-b=2|num-a=4|n=I}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

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 $\boxed{529}$.

See Also

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

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