Answer
$f(t)=\displaystyle \frac{3000}{1+29(2^{1/5})^{-t}}$
It will take about $16$ days.
Work Step by Step
Logistic Model form: $\displaystyle \quad f(t)=\frac{N}{1+Ab^{-t}}$
$N$ is the limiting value, the total available market. $N=3000$
Now (t=0 days), we are given
$f(0)=100 \quad \Rightarrow \quad \left\{\begin{array}{ll}
100 & =\frac{3,000}{1+Ab^{-0}}\\
1+A & =3000/100\\
A & =30-1=29
\end{array}\right.$
Initially doubling sales every 5 days, $b^{5}=2$
(because for small t, f(t) behaves exponentially)
So, $b=2^{1/5}$
The model is $f(t)=\displaystyle \frac{3000}{1+29(2^{1/5})^{-t}}$
We want the $t$ for which $f(t)=700$
$700=\displaystyle \frac{3000}{1+29(2^{1/5})^{-t}}$
$1+29(2)^{-t/5}=\displaystyle \frac{3000}{700}$
$29(2)^{-t/5}=\displaystyle \frac{30}{7}-1$
$(2)^{-t/5}=\displaystyle \frac{23/7}{29}=\frac{23}{203}$
$\displaystyle \log(2)^{-t/5}=\log\frac{23}{203}$
$-\displaystyle \frac{t}{5}\log 2=\log\frac{23}{203}$
$t=-\displaystyle \frac{5\log\frac{23}{203}}{\log 2}\approx 15.708$ days
It will take about $16$ days.
