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

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== Solution 1 ==
 
== Solution 1 ==
 
Substitute <math>y = \frac34x</math> into <math>x^y = y^x</math> and solve.
 
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^{-\frac14} = \frac34</cmath>
 
<cmath>x = \frac{256}{81}</cmath>
 
<cmath>x = \frac{256}{81}</cmath>
Line 27: Line 27:
 
Taking the logarithm base <math>x</math> of both sides, we arrive with:
 
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>
+
<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,
 
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>
 
<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>
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>
 
<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 3 ==
+
== Solution 4 (another version of Solution 3)==
−
<cmath>x^{\frac34x} = (\frac34x)^x</cmath>
+
Taking the logarithm base <math>x</math> of both sides, we arrive with:
−
<cmath>x^{\frac34x} = (\frac34)^x \cdot x^x</cmath>
+
<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>
−
<cmath>x^{-\frac14x} = (\frac34)^x</cmath>
+
Now we proceed by the logarithm rule <math>\log(ab)=\log a + \log b</math>. The equation becomes:
−
<cmath>x^{-\frac14} = \frac34</cmath>
+
<cmath>\log_x \frac{3}{4} + \log_x x = \frac{3}{4}</cmath>
−
<cmath>x = \frac{256}{81}</cmath>
+
<cmath>\Longleftrightarrow \log_x \frac{3}{4} + 1 = \frac{3}{4}</cmath>
−
<cmath>y = \frac34x = \frac{192}{81}</cmath>
+
<cmath>\Longleftrightarrow \log_x \frac{3}{4} = -\frac{1}{4}</cmath>
−
<cmath>x + y = \frac{448}{81}</cmath>
+
<cmath>\Longleftrightarrow x^{-\frac{1}{4}} = \frac{3}{4}</cmath>
−
<cmath>448 + 81 = \boxed{529}</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 ==
 
== See Also ==

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)
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Problem 4
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