Posted by Nycw on Friday, February 1, 2013 at 9:32pm.
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