Answer
$\displaystyle \qquad f(x)=\frac{200}{1+19(2^{-x})}$
Work Step by Step
Goal: $\displaystyle \quad f(x)=\frac{N}{1+Ab^{-x}}$,
where $N$ = limiting value = $200$ (given)
$f(0)=10\Rightarrow\left\{\begin{array}{ll}
10 & =\dfrac{200}{1+Ab^{0}}\\
1+A & =200/10\\
A & =20-1=19
\end{array}\right.\qquad \Rightarrow A=19.$
For small values of x, $f(x)=\displaystyle \frac{N}{1+Ab^{-x}}\approx(\frac{N}{1+A})b^{x}$
(the logistic function grows approximately exponentially with base b.)
$f(x)\displaystyle \approx(\frac{200}{20})b^{x}=10b^{x}$
To double with an increase of x by 1, b must be 2:
$f(x+1)=2f(x)$
$ 10b^{x+1}=2(10b^{x})\qquad$ (divide with $10b^{x})$
$b=2$
Thus, $\displaystyle \qquad f(x)=\frac{200}{1+19(2^{-x})}$
