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?

Answer 1

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