The doubling period of a bacterial population is 10 minutes. At time t = 100 minutes, the bacterial population was 60000

Question

What was the initial population at time t = 0 ?Find the size of the bacterial population after 4 hours.

Answer 1

in 100 minutes it doubled 10 times (100 = 10x10)

60,000 in 100 minutes means

30,000 in 90 minutes

15,000 in 80 minutes

7,500 in 70 minutes

3,750 in 60 minutes

1,875 in 50 minutes

937.5 in 40 minutes

468.75 in 30 minutes

234.375 in 20 minutes

117.1875 in 10 minutes

58.59375 at time t = 0

in 100 minutes there are 10 doubling times, 2^10 = 1024

x(1024) = 60,000

x = 60,000/1024 = 58.59375 = amount at time t=0

in 4 hours = 4 x 60 = 240 minutes

60,000 in 100 minutes

120,000 in 110 minutes

240,000 in 120

480,000 in 130

960K in 140

1,920K in 150

3.840K in 160

7,680K in 170

15,360K in 180

30,720K in 190

61,440K in 200

122,880K in 210

245,760K in 220

491,520K in 230

983,040K in 240 = 4 hours

983,040,000 in 4 hours

60,000 = 60K in 100 minutes

240= 100+140 = 100 = 14(10).

4 hours means 14 doubling times since 100 minutes

2^14 = 16384

60,000 x 2^14 = 98304 x 10^4 = 983,040,000