The Logistic Function (Population Growth)

The population growth rate is given by $P'(t) = \frac{ABCe^{-Ct}}{(1+Be^{-Ct})^2}$ and the population acceleration rate is given by $P''(t) = \frac{AB^2C^2e^{-2Ct}-ABC^2e^{-Ct}}{(1+Be^{-Ct})^3}$ Since $\lim_{t\rightarrow\infty} P(t) = A$ , putting $P(t) = \frac{4A}{5}$ , we get $t_2 = 2\lg 2$ and putting $P''(t) \leq 0$ gives us $t_1 \geq \lg 2$

If you know some properties of the logistics function, one is that it's derivative starts decreasing when the function is at 50% of limit. That's the key here. You get ln(2) for this and 2ln(2) for the 80% case. Hence 3.

I don't know why people are trying to solve this ;)

Look at the graph guys. Graph is created for a reason, to simplify our lives. I didn't even solve the equation. Since it says, when does the growth starts to decrease and reaches 80%, the slope of the graph will be tending towards zero (no or little rise on y-axis, compared to the run on the x-axis) :P

The nearest possible corresponding x value at the specific point is given by x = 3, that's how we get it

for t1,it is given p'(t) is declining.we assume f(x)=p'(t),as f(x )is decreasing at t1,so f(x ) has a maximum value at t1.make f'(x)=0 or p"(t)=0.it will give thr maxima t=ln2,whic is actually,t1.therefore we get t1=ln 2.For t2,proceed as Purushottam Kar's method given below..

T=3