Finding roots of a nonlinear equation I plotted the functions $y=2x$ and $y=\ln(1+x)$ and they have two intersecting points. So the equation $2x=\ln(1+x)$ is supposed to have two roots. One of them is obviously $x=0$. How can I get the other one?
 A: If $x = -t/2-1$, the equation says $\ln(-t/2) = -t - 2$. Exponentiating,
$-t/2 = \exp(-t-2)$, i.e. $t \exp(t) = -2 \exp(-2)$.  Thus $t$ is one of the branches of
$W(-2 \exp(-2))$, where $W$ is the Lambert W function.  Since $-1/e < -2 \exp(-2) < 0$, there are two real branches, the principal branch and the $-1$ branch.  In this case the $-1$ branch is $-2$, leading to $x=0$,  the principal branch is approximately $-0.4063757400$, leading to $x = -t/2-1 \approx -0.7968121301$.
A: There is no analytic solution to this equation in terms of elementary functions. Yet, good approximations can still be obtained.
For example, let the root be $r=-1+a$ and expand the function 
$$f(x) = e^{2x} - x -1$$
around $x=-1$ to get,
$$f(r) = 0 \approx f(-1) +f'(-1)a + \frac 12 f''(-1) a^2\tag{1}$$
Evaluate,
$$f(-1) = e^{-2}, \>\>\> f'(-1) = 2e^{-2}-1, \>\>\> f''(-1) = 4e^{-2}$$
Plug above values into (1) to obtain the approximate root,
$$r \approx -1 + \frac 14\left( e^2 -2 - \sqrt{e^4-4e^2-4}\right) = -0.799$$
