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While thinking about the Lambert $W$ function I had to consider

Solving $(x+y) \exp(x+y) = x \exp(x)$ for $y$.

This is what I arrived at:

(for $x$ and $y$ not zero)

$(x+y) \exp(x+y) = x \exp(x)$

$x\exp(x+y) + y \exp(x+y) = x \exp(x)$

$\exp(y) + y/x \exp(y) = 1$

$y/x \exp(y) = 1 - \exp(y)$

$y/x = (1-\exp(y))/\exp(y)$

$x/y = \exp(y)/(1-\exp(y))$

$x = y\exp(y)/(1-\exp(y))$

$1/x = 1/y\exp(y) -1/y$

And then I got stuck.

Can we solve for $y$ by using Lambert $W$ function?

Or how about an expression with integrals?

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    To solve $exp(a)=exp(a+b)$ we can do for $a$ not $0$ : $f_1(a) = ln(-exp(a))$ Then $f_1(f_1(a)) = a + 2 \pi i $ which is a good answer. Likewise let $f(a) = W(-a e^a)$ , then $f(f(a)) = a + b$ wich is probably a solution to $(a+b) e^{a+b} = a e^a$ for $a$ and $b$ not $0$. Maybe I should post this as the answer. However I still do not know if $f(f(a))$ can be simplified or written as an integral ?2012-09-20

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