x^y=y^x

x^x=y^(9y)

I need this in next few hours please help :(

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- Jan 23rd 2008, 03:11 PMAHDDMSolve System
x^y=y^x

x^x=y^(9y)

I need this in next few hours please help :( - Jan 23rd 2008, 03:38 PMmr fantastic
- Jan 23rd 2008, 04:02 PMAHDDM
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 - Jan 23rd 2008, 04:53 PMmr fantastic
$\displaystyle x^2 = 9y^2 \therefore x = \pm 3y$.

Case 1: x = 3y.

$\displaystyle (3y)^{3y} = y^{9y} \Rightarrow 3^{3y} y^{3y} = (y^{3y})^2 \Rightarrow y^{3y} \left( 3^{3y} - y^{3y} \right) = 0 \Rightarrow 3^{3y} - y^{3y} = 0 \Rightarrow 3^{3y} = y^{3y}$. So one obvious solution is $\displaystyle y = 3 \Rightarrow x = 9$ ....... - Jan 23rd 2008, 05:08 PMAHDDM
That looks all right to me but I just plugged in to check it doesnt come out

- Jan 23rd 2008, 05:44 PMmr fantastic
You're right. I should have remembered that the only integer solutions of $\displaystyle x^y = y^x$ are (2, 4) and (4, 2) .....

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

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

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

An*approximate*solution is $\displaystyle y = 1.73205 \Rightarrow x = 5.19615$. 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?