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December 18, 2014

December 18, 2014

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

- calculus -
**Reiny**, Friday, February 1, 2013 at 10:05pmWe 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

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