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

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