calculus
posted by Nycw on .
The count in a bacteria culture was 600 after 10 minutes and 1100 after 30 minutes. What was the initial size of the culture?Find the doubling period. Find the population after 80 minutes. When will the population reach 13000?

We can use any positive base we want, but since they are talking about doubling period, let's use
number = a (2)^(t/k) , where k will be the doubling period and t is in minutes, and a is the initial number
given: when t = 10 , number = 600
600 = a(2)^(10/k)  #1
when t= 30 , number = 1100
110 = a(2)^(30/k)  #2
divide #1 by #2
600/1100 = (a(2)^(10/k)) / (a(2)^(30/k))
6/11 = 2^(10/k  30/k)
6/11 = 2^(20/k)
take log of both sides
log 6  log11 = 20/k log2 , using the rules of logs
20/k = (log6  log11)/log2 = .874469...
k = 20/.874469 ..
= 22.871
So the doubling period is 22.871 minutes
we have to find the initial a
when t=10 , number = 600 , k= 22.871
600 = a(2)^(10/22.871)
a = 443
so amount = 443 (2)^(t/22.871)
when t = 80
number = 443 (2)^(80/22.871) = 50005 or appr 5000
when is number = 13000 ?
13000 = 443 2^(t/22.871)
29.34537.. = 2^(t/22.871)
log 29.34537 = (t/22.871 log2
t/22.871 = log 29.34../log2 = 4.87506..
t = 22.871(4.875..) = 111.4975 min
= appr 111.5 minutes
let's check our 5000 answer
t=0 , number = 443
t = appr 23 minutes , number = appr 886
t = appr 46 minutes, number = appr 1772
t = 69 minutes , number = appr 3544
t = 92 minutes , number = appr 7088
we had t  80 for a number of 5000
makes sense