The count in a bacteria culture was 500 after 10 minutes and 1300 after 40 minutes.

Question

The count in a bacteria culture was 500 after 10 minutes and 1300 after 40 minutes. Assuming the count grows exponentially. The initial size is 363.43, and the doubling period is 21.87min.

1. Find the population after 95 minutes.   
2. When will the population reach 13000. 

Answer 1

Po=363.43e^0r at time t=0, r=% rate of change per minute

500 = 363.43e^10r at time t=10 minutes

726.86=363.43e^21.87r at time t=21.87 minutes

2 = e^21.87r

1300 = Poe^40t at time t= 40 minutes

P =Poe^95r = population in 95 minutes = how many?

13,000 = Poe^rt = population in t minutes, what is t?

calculate r, plug it into above to find P and t

500/363.43 = e^10r

ln(500/363.43) = 10r

r = (.1)ln(500/363.43)=about .0319 = about 3.19% per minute

2 = e^21.87r

ln2 = 21.87r

r = (ln2)/21.87 = about .03169 = about 3.17% per minute

1300 = 363.43e^40r

r = (ln(1300/363.43)/40 = about .03186 = about 3.19% per minute

the given information is slightly contradictory or inconsistent

go with r= .0319, e= 2.718181818...

use a calculator with natural log and exponential functions

P = 363.43e^95(.0319) = population after 95 minutes = about 7525.76 = about 7526

population = 13,000 when

13,000 = 363.43e^.0319t solve for t

t = (ln(13000/363.43))/.0319 = about 112.13535 = about 112 minutes

Answer 2

Since the doubling time is given, the other growth info isn't needed.

P(t) = P(0)* e ^kt where t is time and k is a constant.

Use the doubling time to find k.

2 = e^k*21.87

ln2 = k*21.87

ln2 / 21.87 = k = about .031694

Population after 95 minutes:

P(95) = 363.43*e^((ln2 / 21.87)*95) = about 7379.89 minutes

13000 = 363.43*e^((ln2 / 21.87)t)

ln 13000 / 363.43 = (ln2 / 21.87)*t

ln(13000/363.43)/(ln2 / 21.87) = t = about 112.86 minutes