The doubling period of a bacterial population is 20 minutes. At time t =120 minutes, the bacterial population was 80000.

Question

what was the initial population at time t= 0?

find the size of the bacterial population after 4 hours

Answer 1

80000 = a₀(2)^120/20

80000 = a₀(2)⁶

80000 = a₀(64)

1250 = a₀

a₄ = 1250(2)^240/20

Can you calsulate and answer?

Answer 2

2 = e^20r where r = rate of change per minute

80000= (Po)e^120r where Po = initial amount at time t=0

80,000=1,250e^120r

e^120r = 80,000/1250

120r = ln(8000/125)=ln64

r = 4.159/120 = about .0347

A = 1250e^.0347t

it doubles every 20 minutes

80,000 at t=120

40,000 at t=100

20,000 at t=80

10,000 at t=60

5,000 at t = 40

2,500 at t=20

1,250 at t= 0

initial amount of bacteria = 1,250

A=1250e^.0347(0) = 1250(1) = 1,250

4 hours = 4(60) = 240 minutes

80,000 at 120 minutes

160,000 at 140 minutes

320,000 at 160 minutes

640,000 at 180 minutes

1,280,000 at 200 minutes

2,560,000 at 220 minutes

5,120,000 at 240 minutes = 4 hours

A=1250e^.0347(240)

= close to 5 million in 4 hours, rounding errors for r make it slightly more than 5,120,000