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>. | 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>. | ||
Revision as of 16:02, 9 August 2018
Contents
Problem
Suppose that 
and 
. The quantity 
can be expressed as a rational number 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
We solve in general using 
instead of 
. Substituting 
, we have:
![\[x^{cx} = (cx)^x \Longrightarrow (x^x)^c = c^x\cdot x^x\]](http://latex.artofproblemsolving.com/6/d/8/6d806bc47a9f272be605a24af932316a92ed65fd.png)
Dividing by 
, we get 
.
Taking the 
th root, 
, or 
.
In the case 
, 
, 
, 
, yielding an answer of 
.
Solution 2
Taking the logarithm base of both sides, we arrive with:
![\[y = log_x y^x \Longrightarrow \frac{y}{x} = log_x y = log_x \frac{3}{4}x = \frac{3}{4}\]](http://latex.artofproblemsolving.com/d/b/3/db3df9c38b9ecedfee44276f583cefbd96624ccd.png)
Where the last two simplifications were made since 
. Then,
![\[x^{\frac{3}{4}} = \frac{3}{4}x \Longrightarrow x^{\frac{1}{4}} = \frac{4}{3} \Longrightarrow x = \left(\frac{4}{3}\right)^4\]](http://latex.artofproblemsolving.com/c/b/0/cb0f2145072150b5a6cefb8eeed0ca8aadc30efe.png)
Then, 
, and thus:
![\[x+y = \left(\frac{4}{3}\right)^3 \left(\frac{4}{3} + 1 \right) = \frac{448}{81} \Longrightarrow 448 + 81 = \boxed{529}\]](http://latex.artofproblemsolving.com/d/4/3/d43be6aefd097e415bc26154289085ca723e9c25.png)
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
