How to algebraically derive the inverse of $x + e^x = y$?

Question

Okay, so here's an approach I took:

$$ x + e^x = y $$ $$ e^{[x + e^x]} = e^y $$ $$ e^x e^{e^x} = e^y $$ $$ e^x = W(e^y) $$ $$ x = \ln{W(e^y)} $$ Where $W(z)$ is Lambert W function.

This works, but if I feed the initial problem into Wolfram Alpha, it gives me a different result: $$ x = y - W(e^y) $$

This solution looks cleaner and works better for the purposes of the problem I'm trying to solve. However since I don't have much experience doing math with the W function and utilizing its properties, I failed to figure out, how I derive it myself. Please, may someone explain to me, how?

Answer 1

One has $W(e^y)e^{W(e^y)} = e^y$ by definition of the Lambert function, hence $W(e^y) = e^{y-W(e^y)}$ and finally $x = \ln W(e^y) = y - W(e^y)$.

Answer 2

1)

$$x+e^x=y$$

We see, the elementary function term on the left-hand side of the equation is a polynomial without a univariate factor and in dependence of two $\mathbb{C}$-algebraically independent monomials ($x,e^x$). Therefore we don't know how to solve the equation for $x$ by rearranging for $x$ by applying only finite numbers of elementary functions that we can read from the equation. We therefore cannot see if the elementary function on the left-hand side has a partial inverse that is an elementary function.

Furthermore, for algebraic $y$, the equation is an irreducible algebraic equation of $x$ and $e^x$. The theorem in Lin 1983 and the theorem in Chow 1999 state that such kind of equations don't have solutions in the elementary numbers and in the explicit elementary numbers respectively, if Schanuel's conjecture is true.

2)

Lambert W is the inverse relation of the function $x\mapsto xe^x$.

To see if our equation is solvable in terms of Lambert W, we try to rearrange the equation into the product form that's required for Lambert W.

This is a longer way, but without any tricks.

A sum equation can be rearranged to a product equation in each case.

$$x+e^x=y$$ $$e^x=y-x$$ $$1=(y-x)e^{-x}$$ $$(y-x)e^{-x}=1$$ $$(y-x)e^{y-x}=e^{y}$$ $$y-x= W(e^y)$$ $$x=y-W(e^y)$$

Answer 3

$x=y-W(e^{y})$

The general equation:

$e^x +x=y$

Subtract x from both sides

$e^{x}=y-x$

Multiply both by $e^{-x}$

$1=(y-x)e^{-x}$

Multiply both by $e^{y}$

$(y-x)e^{y-x}=e^{y}$

Taking Lambert W of both get:

$(y-x)=W(e^{y})$

Therefore:

$x=y-W(e^{y})$