Rewriting logistic function into standard form My book defines the standard form of a logistic function as $f(x) = A \cdot \frac{1 + B \cdot e^{-\frac{x}{a}}}{1 + C \cdot e^{-\frac{x}{a}}}$, while wikipedia defines it as $f(x) = \frac{L}{1 + e^{-k(x-x_0)}}$. Can anyone show how to rewrite this logistic function into both of these standard forms $$P(t) = \frac{K \cdot P_0 \cdot e^{r\cdot t}}{K + P_0 \cdot (e^{r \cdot t}-1)}$$?
 A: For the first form ...
$$\begin{align}
P(t) &= K\cdot \frac{P_0e^{rt}}{K + P_0 \cdot (e^{rt}-1)} \tag{1}\\[4pt]
&= K\cdot\frac{P_0e^{rt}+ 0}{P_0 e^{rt}+(K-P_0)}\cdot\frac{1/(P_0e^{rt})}{1/(P_0e^{rt})} \tag{2}\\[4pt]
&= K\cdot\frac{1+ 0e^{-rt}}{1+\frac{K-P_0}{P_0}e^{-rt}} \tag{3}\\[4pt]
&= K\cdot \frac{1+Be^{-t/a}}{1+Ce^{-t/a}} \qquad B:=0, C:=\frac{K-P_0}{P_0}, a := \frac1r \tag{4}
\end{align}$$
For the second form, starting from $(3)$ above ...
$$\begin{align}
P(t) &= \frac{K}{1 + \frac{K-P_0}{P_0}e^{-rt}} \tag{5}\\[4pt]
&= \frac{K}{1 + e^se^{-rt}} \qquad s := \log\frac{K-P_0}{P_0} \tag{6}\\[4pt]
&= \frac{K}{1 + e^{-r(t-t_0)}} \qquad t_0 := \frac{s}{r} = \frac1r\log\frac{K-P_0}{P_0} \tag{7}
\end{align}$$
A: As it follows from the wikipedia link, the standard logistic function is
$$f(x) = \dfrac1{1+e^{-x}} = \dfrac{e^x}{e^x+1}.\tag1$$ 
At the same time, any other function can be considered as the generalized logistic function, if can be shown parameters, which transforms it to the form $(1).$
In the first case, these parameters are $A=1,\ B=0,\ C=1,\ a=1.$
In the second case, as it is shown in the link, $k=1,\ x^\,_0 = 0,\ L=1.$
In the third case, $K=1,\ P_0 = \dfrac12.$
Therefore, all functions from OP are the different generalizations of the basic function $(1).$
