The doubling period of a bacterial population is 20 minutes.

Question

At time t=120 minutes, the bacterial population was 60000. What was the initial population at time t=0 ? Find the size of the bacterial population after 4 hours

Answer 1

after 120 minutes population = 60K

double time = 20 minutes

how much at time t= 0??

P(t) = Po(e^rt)

Po = 60K/e^120r)

60k after doubling 6 times = 120/20

30K time t= 5

15K t=4

7.5K t=3

3.75K t= 2

1.875K t= 1

.9375K t = 0

937.5 initially

after 4 hours = 240 minutes = 240/20 = 12 doubling periods

60K for 6

120K for 7

240 for 8

480 for 9

960 for 10

1920 for 11

3840K for 12

3 million, 840 thousand after 4 hours

Answer 2

A(t) = Pe^rt , where t is the number of minutes

2P = Pe^20r and 60000 = Pe^120r

From the first equation, e^20r = 2.

lne^20r = ln2

20r = ln2

r = ln2/20 = 0.034657359

60000 = Pe^{0.034657359(120)}

P ≈ 60000/64 = 937.5 = population when t = 0

So, population in 4 hours = A(240) = 937.5e^{0.034657359(240)} ≈ 3,840,000

Answer 3