There are a few things I say over and over again on math.SE, and one of them is that the cleanest definition of the exponential function (on either $\mathbb{R}$ or $\mathbb{C}$, or more generally even) is that it's the unique function $f : \mathbb{C} \to \mathbb{C}$ (or $\mathbb{R} \to \mathbb{R}$) satisfying
- $f(0) = 1$, and
- $f'(x) = f(x)$.
Note that this definition makes no explicit reference to $e$. Every other property of the exponential function falls easily out of this definition together with the existence and uniqueness theorems for solutions to ODEs. For example, by the chain rule
$$\frac{d}{dz} \exp(z + w) = \exp(z + w)$$
hence $\exp(z + w)$ is also a solution to the above ODE but with initial condition $\exp(w)$. But so is $\exp(z) \exp(w)$. Hence the two are equal by the uniqueness theorem.
Similarly we get continuity at every point and the usual power series expansion. The limit
$$\exp(z) = \lim_{n \to \infty} \left( 1 + \frac{z}{n} \right)^n$$
then falls out of applying the Euler method with step size $\frac{z}{n}$ to approximate solutions to this ODE. (It can also be formally justified by differentiating with respect to $z$ but this requires some thought about exchanging the derivative and the limit.)
This allows us to give a clean definition of $e$ as just being the value $\exp(1)$ (another thing I say over and over again on math.SE is that $e$ is not important, $\exp(z)$ is important and $e$ just happens to be its value at $z = 1$), and a clean definition of $\pi$: with $\exp(z)$ defined as above, $\pi$ is the smallest positive real such that $\exp(2 \pi i) = 1$, or in other words it's half the period of $\exp(it)$. Note that by the chain rule
$$\frac{d}{dt} \exp(it) = i \exp(it)$$
so $\exp(it)$ is a solution to the ODE $f(0) = 1, f'(t) = i f(t)$ for a function $f : \mathbb{R} \to \mathbb{C}$. But $\cos t + i \sin t$ is also such a solution. So by the uniqueness theorem we recover Euler's formula
$$\exp(it) = \cos t + i \sin t.$$
This requires that we know what the trigonometric functions are in advance, but we can actually invent them this way instead. Additivity gives $\exp(it) \exp(-it) = \exp(0) = 1$, but we also have
$$\frac{d}{dt} \exp(it) \overline{\exp(it)} = 0$$
from which it follows that $\exp(-it) = \overline{\exp(it)}$ and that $\| \exp(it) \| = 1$ is a constant. So $\exp(it) = c(t) + i s(t)$ satisfies
$$c(t)^2 + s(t)^2 = 1$$
$$c(-t) = c(t), s(-t) = - s(t)$$
$$c(t_1 + t_2) = c(t_1) c(t_2) - s(t_1) s(t_2)$$
$$s(t_1 + t_2) = c(t_1) s(t_2) + s(t_1) c(t_2)$$
and we're well on our way to rediscovering trigonometry. These identities can be used to show that $\exp(it)$ is periodic by showing that it not only lies on the unit circle but travels on it with constant velocity (this basically follows from additivity).
The same uniqueness idea applied to the trigonometric functions tells us that $(\cos t, \sin t)$ is the unique pair of functions satisfying
- $c(0) = 1, s(0) = 0$, and
- $c'(t) = -s(t), s'(t) = c(t).$
Every other trigonometric identity is a consequence of these. This one may be a little less intuitive but it says that the vector $\left[ \begin{array}{cc} c'(t) \\ s'(t) \end{array} \right]$ is a $90^{\circ}$ rotation of, and in particular orthogonal to, the vector $\left[ \begin{array}{cc} c(t) \\ s(t) \end{array} \right]$, which e.g. after differentiating a second time, exactly describes a particle under the influence of a constant centripetal force.
To make the story short, any equation which can write or rewrite
$$a +b x +c \log(d x+e)=0$$ has explicit solutions in terms of the standard Lambert function.
$$x=-\frac{e}{d}+\frac{c }{b}\,W(k) \qquad \text{where} \qquad k=\frac b {cd}\,\exp\Big[\frac{b e-a d}{c d}\Big]$$
If, on the search bar of the site, you just type Lambert, you will find $3569$ entries.
Best Answer
One of the differences between a logistic function and the arctan is that the logistic function approaches its asymptotes exponentially, i.e. (if $k > 0$) $$\eqalign{f(x) \sim L + B - L \exp(k x_0) \exp(-k x) & \ \text{as $x \to +\infty$}\cr f(x) \sim B + L \exp(-k x_0) \exp(k x) & \ \text{as $x \to -\infty$}}$$ while the arctan approaches its asymptotes much more slowly, like $x^{-1}$: $$ \eqalign{\arctan(x) \sim \frac{\pi}{2} - \frac{1}{x} & \ \text{as $x \to +\infty$}\cr \arctan(x) \sim -\frac{\pi}{2} - \frac{1}{x} & \ \text{as $x \to -\infty$}\cr}$$