# Solve System

• January 23rd 2008, 04:11 PM
AHDDM
Solve System
x^y=y^x
x^x=y^(9y)

• January 23rd 2008, 04:38 PM
mr fantastic
Quote:

Originally Posted by AHDDM
x^y=y^x
x^x=y^(9y)

$(x^y)^x = (y^x)^x \Rightarrow x^{yx} = y^{x^2}$.

$(x^x)^y = (y^{9y})^y \Rightarrow x^{xy} = y^{9y^2}$.

Therefore $y^{x^2} = y^{9y^2} \Rightarrow x^2 = 9y^2$.

Therefore ........
• January 23rd 2008, 05:02 PM
AHDDM
x^2=9y^2

y=x/3

x^x=y^(9*x/3)
x^x=y^(3x)

I got this so far<

When I use x=3y and plugin to first equation

I get to 3y^y=y^(3y)

I did it and i got x=0 and y=0
• January 23rd 2008, 05:53 PM
mr fantastic
Quote:

Originally Posted by AHDDM
x^2=9y^2

y=x/3

x^x=y^(9*x/3)
x^x=y^(3x)

I got this so far<

When I use x=3y and plugin to first equation

I get to 3y^y=y^(3y)

$x^2 = 9y^2 \therefore x = \pm 3y$.

Case 1: x = 3y.

$(3y)^{3y} = y^{9y} \Rightarrow 3^{3y} y^{3y} = (y^{3y})^2 \Rightarrow y^{3y} \left( 3^{3y} - y^{3y} \right) = 0 \Rightarrow 3^{3y} - y^{3y} = 0 \Rightarrow 3^{3y} = y^{3y}$. So one obvious solution is $y = 3 \Rightarrow x = 9$ .......
• January 23rd 2008, 06:08 PM
AHDDM
That looks all right to me but I just plugged in to check it doesnt come out
• January 23rd 2008, 06:44 PM
mr fantastic
Quote:

Originally Posted by AHDDM
That looks all right to me but I just plugged in to check it doesnt come out

You're right. I should have remembered that the only integer solutions of $x^y = y^x$ are (2, 4) and (4, 2) .....

So (9, 3) is a spurious solution ......

If you sub x = 3y into $x^y = y^x$ you get:

$(3y)^y = y^{3y} \Rightarrow 3^y y^y = (y^y)^3$ $\Rightarrow y^y \left( 3^y - (y^y)^2 \right) = 0$ $\Rightarrow 3^y - y^{2y} = 0$.

An approximate solution is $y = 1.73205 \Rightarrow x = 5.19615$. This solution does seem to work ..... To express this solution for y in exact form, I think you'd need to use the Lambert W-function ......

I haven't looked at the x = -3y case yet, but there will probably be similar if not worse difficulty in store ......

Are you expected to get exact solutions?