How does logistic growth rate coincide with the slope of the line in the exponential phase of the growth? For a logistic function $$f(x) = \frac{L}{1 + e^{-k(x - x_0)}},$$ people call $k$ the logistic growth rate. Now, I have encountered this statement: In the log scale the logistic growth rate coincides with the slope of the line in the exponential phase of the growth. If this is correct, how can one prove it?

Let's do the calculations to see what the answers are.
By changing the units of measurement of $x$ to the origin $x_0$ we may assume $x_0=0$ (to simplify the work and the notation) and--therefore--the middle of the curve is at $x=0.$  Thus
$$\frac{\mathrm{d}}{\mathrm{d}x} \log(f(x)) = \frac{\mathrm{d}}{\mathrm{d}x}\left(\log(L) - \log(1 + e^{-kx})\right) = \frac{ke^{-kx}}{1 + e^{-kx}}.$$
In the middle at $x=0$ this simplifies to $k/2.$
However, if we take "exponential phase of the growth" to refer to the region where $x\ll x_0,$ its slope has a limiting value
$$\lim_{x\to-\infty} \frac{ke^{-kx}}{1 + e^{-kx}} = \lim_{x\to-\infty} \frac{k}{1 + e^{kx}} = k.$$
The derivative of $f$ itself is obtained by multiplying the logarithmic derivative by $f,$ giving
$$\frac{\mathrm{d}}{\mathrm{d}x} f(x) = \frac{ke^{-kx}}{1 + e^{-kx}}\frac{L}{1 + e^{-kx}}.$$
In the middle at $x=0$ this simplifies to $kL/4.$
This leads to three useful rules, illustrated below:


*

*The limiting slope at the left of the graph of $\log f$ is $k.$

*The middle slope of the graph of $\log f$ is half the rate, $k/2.$

*When $L=4,$ the middle slope of the graph of $f$ equals the rate $k.$

The special role of $L=4$ is revealed by studying the folded log and its inverse.

The left hand curve suggests the limiting log slope is achieved closely for $x \ll -3/k.$  The right hand curve indicates $kL/4$ is a good approximation of the slope for $|x| \lt 1/k.$  These observations provide guidance for estimating $k$ from data, especially when only a rough "eyeball" estimate might be needed.
