The population of a culture of bacteria is modeled by the logistic equation P (t) = frac{14,250}{1 + 29e – 0.62t} To the nearest tenth, how many days will it take the culture to reach 75% of it’s carrying capacity? What is the carrying capacity? What are the virtues of logistic model?

Step1 $P(t)=\frac{14,250}{1+29e^{(}-0\ast 62t)}$
$Ast\Rightarrow \mathrm{\infty },e^{-0\ast 62t}\Rightarrow 0$
$\therefore P(t)\Rightarrow 14,250$.

This is the carrying capacity of the bacten`al culture.

To determine the time it take to reach 75% of thr carlying capacity, we write $P(t)=\frac{3}{4}(14,250).$

Step 2

$\Rightarrow \text{⧸}14,2501+29e^{(-0\ast 62t)}=\frac{3}{4}(\text{⧸}14,250)$
$\Rightarrow 1+29e^{-0\ast 62t}=\frac{3}{4}$
$\Rightarrow e^{(-0\ast 62t)}=\frac{1}{87}$
$\Rightarrow t=\frac{1}{0\ast 62}In(87)=7\cdot 2$ Hense it takes the bacterial Popylation $7\cdot 2$ days to reach 75% of the calling capacity.

Step 3 The growth of any population is curtailed by the amount of resources available. This is better modeled by a logistic S curve rather than an exponential growth model which only works in ideal situations and not for modeling real life situations.