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2010 AIME I Problems/Problem 3: Difference between revisions

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<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 = (\frac{4}{3})^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>
Then, <math>y = (\frac{4}{3})^3</math>, and thus:
Then, <math>y = \left(\frac{4}{3}\right)^3</math>, and thus:
<center> <cmath>x+y = (\frac{4}{3})^3 (\frac{4}{3} + 1) = \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>


== See Also ==
== See Also ==

Revision as of 20:17, 11 March 2017

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:

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

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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