Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 ? – mick Sep 20 '12 at 21:24

The solution of $ (x+y) \exp(x+y) = x \exp(x) $ is given in terms of the Lambert W function

Let $z=x+y$, then we have

$$ z {\rm e}^{z} = x {\rm e}^{x} \Rightarrow z = { W} \left( x{{\rm e}^{x}} \right) \Rightarrow y = -x + { W} \left( x{{\rm e}^{x}} \right) \,. $$

Added: Based on the comment by Robert, here are the graphs of $ y = -x + { W_0} \left( x{{\rm e}^{x}} \right) $ and $ y = -x + { W_{-1}} \left( x{{\rm e}^{x}} \right) $

enter image description here

enter image description here

share|cite|improve this answer
Hmm but if $W(xe^x) = x$ , then $y = -x + x = 0$ ? – mick Sep 20 '12 at 21:04
I got another ' solution ' , see my other comment. – mick Sep 20 '12 at 21:27
There are different branches of the Lambert W function. In Maple's numbering, $W_0(x e^x) - x = 0$ for $x \ge -1$, while $W_{-1}(x e^x) - x = 0$ for $x \ge -1$. For $x \le -1$, $W_0(x e^x) - y$ is the nonzero real solution of $(x+y) e^{x+y} = x e^x$. For $-1 \le x < 0$, $W_{-1}(x e^x) - x$ is the nonzero real solution. – Robert Israel Dec 25 '12 at 5:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.