How to solve $e^x=kx + 1$ when $k > 1$?

Question

It's obvious that $x=0$ is one of the roots. According to the graphs of $e^x$ and $kx + 1$, there's another root $x_1 > 0$ when $k > 1$. Is there a way to represent it numerically.

Answer 1

If $t = -x - 1/k$, we have $$t e^t = (-x - 1/k) e^{-x} e^{-1/k} = - e^{-1/k}/k $$ The solutions of this are $t = W(-e^{-1/k}/k)$, i.e. $$x = - W(-e^{-1/k}/k) - 1/k$$ where $W$ is one of the branches of the Lambert W function. If $k > 1$, $-e^{-1/k}/k \in (-1/e,0)$, and there are two real branches: the $0$ branch (which gives you $x=0$), and the $-1$ branch, which gives you the solution you want.

Answer 2

$$e^x=kx+1$$ $$x=\log(1+kx)$$ so $$x=\log(1+k\log(1+k\log(1+k\log(1+....))))$$