calculus

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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?

  • calculus -

    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

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