Solving for $x$ in an exponential equation Say we the following equation
$$F(x) = \frac{\exp(a+bx)}{1 + \exp(a+bx)}$$
Now we set $x=0$ and we want to solve for $a$ as a function of $F_0$.
So that, we have:
$$F_0 = \frac{\exp(a)}{1 + \exp(a)}$$
Can someone please walk me though the logarithmic transformation on how we achieve this end result
$$a = \ln\left(\frac{F_0}{1-F_0}\right)$$
 A: Put $y = e^{a+bx}, c = F(0)  \to c = \dfrac{y}{1+y} \to y = c + cy \to y - cy = c \to (1-c)y = c \to y = \dfrac{c}{1-c} \to e^{a+bx} = \dfrac{c}{1-c} \to a+bx = \ln\left(\dfrac{c}{1-c}\right) = d \to x = \dfrac{d-a}{b}=\dfrac{\ln \left(\dfrac{F(0)}{1-F(0)}\right)-a}{b}$
Note: Solve for $a$ is quicker and you can see from above that I want to solve for $x$ which is a bit different but equally good.
A: $$F_0=\frac{\exp(a)}{1+\exp(a)}$$
This leads to:
$$(1+\exp(a))F_0=\exp(a)$$
Thus:
$$F_0=\exp(a)(1-F_0)$$
Dividing by $1-F_0$ and taking the log, you get the result.
A: Hint
Since you look for $a$ such that $$F_0 = \frac{e^a}{1+e^a}$$ define first $x=e^a$ so the equation becomes $$F_0 = \frac{x}{1+x}$$ Solve for $x$ and take the logarithms.
A: $$F_0=\frac{e^a}{1+e^a}$$
$$F_0+e^a \cdot F_0=e^a$$
$$F_0=e^a-e^a \cdot F_0$$
$$F_0=(1-F_0) \cdot e^a$$
$$e^a=\frac{F_0}{1-F_0}$$
$$a=\ln\left(\frac{F_0}{1-F_0}\right)$$
A: Notice, we have $$F(x)=\frac{exp(a+bx)}{1+exp(a+bx)}=\frac{e^{a+bx}}{1+e^{a+bx}}$$ Setting  $x=0$, we get  $$F_0=\frac{e^a}{1+e^a}$$
$$1-F_0=1-\frac{e^a}{1+e^a}$$  $$1-F_0=\frac{1+e^a-e^a}{1+e^a}$$
$$1-F_0=\frac{1}{1+e^a}$$
 $$1+e^a=\frac{1}{1-F_0}$$
$$e^a=\frac{1}{1-F_0}-1=\frac{1-1+F_0}{1-F_0}$$$$e^a=\frac{F_0}{1-F_0}$$
Taking log on both the sides, we get $$\ln e^a=\ln\left(\frac{F_0}{1-F_0}\right)$$
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{a=\ln\left(\frac{F_0}{1-F_0}\right)}}$$
A: $f(x)=\Large\frac{e^{(a+bx)}}{1+e^{(a+bx)}}$
when $x=0, f(0)=\Large\frac{e^{(a+0)}}{1+e^{(a+0)}}=\frac{e^a}{1+e^a}$
multiply and divide the RHS by $e^a$
$f(0)=\frac{1}{\large\frac{1}{e^a}+1}\Rightarrow \frac{1}{e^a}+1=\frac{1}{f(0)}$
$\therefore \frac{1}{e^a}=\frac{1}{f(0)}-1=\frac{1-f(0)}{f(0)}$
$\therefore e^a=\frac{f(0)}{1-f(0)}$
$\therefore \ln e^a = \ln \left(\frac{f(0)}{1-f(0)}\right)$
$\ln e^a=a \ln e$ & $\ln e=1$,
$\therefore a= \ln \left(\frac{f(0)}{1-f(0)}\right)$
