Equation 1:
$x+e=e^x$
According to Wolfram alpha : Solution of x $\approx$ -2.6 and 1.4
Equation 2:
$x-e = \ln(x)$
According to wolfram alpha, Solution for x $\approx$ 0.07 and 4.1
How does one solve equations like this? For exact solution they use one W() Function ... What is that?
Thanks a lot!
That $W$ is the Lambert W function. (But I thought Wolfrom didn't use that terminology, instead "product logarithm" or something?)
added
The Lambert W function is inverse to the function $xe^x$. So that means: $xe^x=y$ becomes $x=W(y)$. In some cases there may be two real solutions. And there are infintely many complex solutions. Represented by "branches" of the $W$ function: $\dots, W_{-1}, W_0, W_1, \dots$.
So, let's do the problem $$ x+e = e^x . $$ I rearrange this: $$ (x+e)e^e = e^{x+e} \\ -(x+e)e^{-(x+e)} = -e^{-e} \\ x+e = W\left(-e^{-e}\right) \\ x = e+ W\left(-e^{-e}\right) $$
Maple shows that the two real branches of this are: $$ e+ W_0\left(-e^{-e}\right) \approx 2.64745 \\ e+ W_{-1}\left(-e^{-e}\right) \approx -1.42037 $$