contestada

A common inhabitant of human intestines is the bacterium escherichia coli. a cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. the initial population of a culture is 71 cells. (a) find the relative growth rate. (assume t is measured in hours.)

Respuesta :

underV
You need to use this formula to solve this exercise:  P(t)=Ae^(k.t)

P(t) is the population of bacteria
A is the number of bacteria at the beginning of the experiment: 71
K is the growth rate

20 min= 1/3 hours
P(0)=71
P(1/3)= 71 ⇔ P= 71 x 2 ⇔142 

142= 71 x e^k/3
2= e^k/3
ln 2 = k/3
3ln 2=k
ln(2³)=k
k=ln8


W0lf93
The general formula is Po x 8^T, where Po is the initial population and T is the time in hours. Using the original statement of 71 equaling Po one would get the formula 71 x 8^T, in effect growing logarithmically by 8 every hour, going from 568 at hour one to 4544 by the end of hour two and so forth.