A cell of some bacteria divides into two cells every 40 minu

A cell of some bacteria divides into two cells every 40 minutes. The initial population is 3 bacteria.

(a) Find the size of the population after t hours

(b) Find the size of the population after 2 hours.

(c) When will the population reach 66?

Solution

Part-1)

Bacterial growth follows an exponential growth model that can be represented as:

y = y(0) e^(kt)

where y is the population after t time, y(0) is the initial population, k is a constant specific to the problem, and t is time.

For your problem, assume that time is measured in hours. 40 minutes is 2/3 of an hour, so it corresponds to t = 2/3. We know that after 2/3 of an hour, 3 bacteria become 6 bacteria, so we can write:

y = y(0) e^(kt)
6 = 3e^(k * 2/3)

Solving for k:

2 = e^(2k/3)
ln 2 = 2k/3
3 ln 2 = 2k
3/2 (ln 2) = k

Now that we know k, we can write a general equation for the population growth after t hours.

y = 3e^((3/2) (ln 2) t)

     
Part-2

put t= 2 hours

we get-

y = 3e^((3/2) (ln 2) 2)

y = 3e^((3ln2)

y=3*8

y=24

The size of the population after 2 hours =24

Part-3

put y=66

  66 = 3e^((3/2) (ln 2) t)

22=e^((3/2)(ln2)t)

taking both side logarithm

ln(22)=(3/2)(ln2)(t)

ln(22)=3.09104245336 and ln(2)=0.69314718056

ln(22)/ln(2)=4.45943161864

(4.45943161864)(2/3)=t

  2.97295441243=t (in hours)

The population reach 66 in 2.972 hours