Solutions to $x+e^x=k$

Question

So I am trying to solve $x+e^x=k$ and here is what I have done:

$$x+e^x=k$$ $$e^{x+e^x}=e^k$$ $$e^xe^{e^x}=e^k$$ Now, if we use the lambert W function which has the identity such that if $y=xe^x$ then $x=W(y)$ $$e^x=W(e^k)$$ $$x=\ln(W(e^k))$$ However, when I looked it up on WolframAlpha, they gave the answer $x=k-W(e^k)$

Did I do something wrong here?

Answer 1

I would like to remind you that

$$z = W(z)e^{W(z)}$$

defines $W(z)$.

$$k - W(e^k) = \log(W(e^k))$$

$$e^ke^{ - W(e^k)} = W(e^k)$$

$$e^{k} = W(e^k)e^{ W(e^k)}$$

$$e^k = e^k$$

In other words, you have just discovered that: $$\log({W(z)}) = \log(z) - W(z).$$