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?

If L = 1 and one transforms to the logit (log odds) scale $p(x)$ then the slope is exactly k.  The right hand side of the first equals is by definition of log odds and the second is by algebraic simplification.
$p(x) = log(\frac{f(x)}{1-f(x)}) = k(x - x_0)$
Now
$dp(x)/dx = k$
This is the usual interpretation of k used in logistic regression.
Added
In general, define $p(x) = log(\frac{f(x)}{L-f(x)})$ in which case L cancels, $p(x) = k(x-x_0)$ as above and the above derivative still holds.
