Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 3)
(→‎See Also)
Line 32: Line 32:
 
Then, <math>y = \left(\frac{4}{3}\right)^3</math>, and thus:
 
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>
 
<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 ==
 +
Taking the natural logarithm of both sides, we get:
 +
 +
<center><cmath> \ln {x^y} = \ln {y^x} </cmath></center>
  
 
== See Also ==
 
== See Also ==

Revision as of 08:52, 31 May 2020

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

Taking the natural logarithm of both sides, we get:

\[\ln {x^y} = \ln {y^x}\]

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_3&oldid=123341"