x^y=y^x
x^x=y^(9y)
I need this in next few hours please help
$\displaystyle x^2 = 9y^2 \therefore x = \pm 3y$.
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 $\displaystyle y = 3 \Rightarrow x = 9$ .......
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?