What is the initial population at time t=0?

Question

The doubling period of a bacterial population is 20 minutes. At time t=100 minutes, the bacterial population was 60,000. What was the initial population at time=0?

Find the size of the bacterial population after 4 hours.

Answer 1

2 = e^rt where t = time measured in minutes

2 = e^r20 take the natural log of both sides

ln2 = 20r solve for r

r = ln2/20 = 0.0347 = rate of growth

60,000 = Ae^0.0347(100) = Ae^3.47

A= 60,000/e^3.47 = 60,000/32.137 = 1,867 = original amount of bacteria at time t=0

after 4 hours, t=4x60= 240

P=1867e^0.0347(240) = 1867e^8.328 =(1867)(4138.133) = 7.725,894.319= over 7 2/3 million

at t=100, P = 60,000

every 20 minutes it doubles

in 4 hours that's 240 minutes, another 140 minutes or 140/20 = 7 doubling periods

60,000x(2)^7 = 60,000 x 128 = 7,680,000 in 4 hours