Answer
$f(t)=\displaystyle \frac{100}{1+49(\sqrt{2})^{-t}}$
The average consumption of 50 per year is to be reached during 2004.
Work Step by Step
Logistic Model form: $\displaystyle \quad f(t)=\frac{N}{1+Ab^{-t}}$
$N$ is the limiting value, the saturation estimate. $N=100.$
Now (t=0 years), we are given
$f(0)=2 \quad \Rightarrow \quad \left\{\begin{array}{ll}
2 & =\frac{100}{1+Ab^{-0}}\\
1+A & =50\\
A & =49
\end{array}\right.$
The rate of increase was double in two years, $b^{2}=2$
(because the model behaves exponentially early on).
So, $b=\sqrt{2}$
$f(t)=\displaystyle \frac{100}{1+49(\sqrt{2})^{-t}}$
We want the $t$ for which $f(t)=50$
$50=\displaystyle \frac{100}{1+49(\sqrt{2})^{-t}}$
$1+49(\sqrt{2})^{-t}=100/50$
$49(\sqrt{2})^{-t}=2-1=1$
$(\sqrt{2})^{-t}=1/49$
$\log(\sqrt{2})^{-t}=\log(1/49)$
$-t\log\sqrt{2}=-\log 49$
$t=\displaystyle \frac{\log 49}{\log\sqrt{2}}\approx 11.2219$
So $t\approx 11$, (years from 1993)
The average consumption of 50 per year is to be reached during 2004.
