Solve $ x^y=7$ and $y^x=3$, for $ (x,y)∈\mathbb{R^2}$ with $x\neq y$. Is there any algebraic solution to the following system of equations:
$x^y=7$ and $y^x=3$ for $ (x,y)∈\mathbb{R^2}$ with  $x\neq y$.
The only thing I can do is using a root solving technique (like Newton Rhapson) to find an answer, or a math software: I arrived to find an approximately root: $x=4.64689, y= 1.2667$.
Thanks for your help.
 A: $x^{y} = 7.....(1)$
$y^{x} = 3.....(2)$
taking equation (1)
$x^{y} = 3^{log_{3}7} $
let $log_{3}7 = a$
$x^{y} = 3^{a} $
$x^{y} = (y^{x})^{a} $
$x^{y} = y^{ax} $
let y= ux
$x^{ux} = (ux)^{ax} $
$(x^{u})^{x}= (u^{a}x^{a})^{x} $
$x^{u}= u^{a}x^{a} $
$x^{u-a}= u^{a} $
$x= u^{(\frac{a}{u-a})} $
since $y = ux$
$y= u \times u^{(\frac{a}{u-a})} $
$y= u^{(\frac{u}{u-a})} $
Therefore answers
$x= u^{(\frac{a}{u-a})}, y= u^{(\frac{u}{u-a})} $
Now I`m painfully aware that this does not actually produce the right answer, unless you use
$u = \frac{y}{x} = \frac{1.2667 }{4.64689} = 0.272590916$ which is cheating, I could fill this whole page with the numerous ways in which I tried to solve this problem.I guess it was just painful to let it go without posting something. This attempt was inspired by the proof for $x^{y} = y^{x} $ where the $y=ux$ works by removing the powers of x from both sides. I conclude that there is no way of proving this algebraically but maybe I`m just as pathetic as my low self-confidence would suggest.
