Post a New Question

Calculus

posted by .

a populatioin of bacteria has an initial size of 500. the population doubles every 7 hours.
a) how many bacteria will there be after 10 hours
b) when will there be 3000 bacter?

I keep getting answers that just don't make sense. Can someone do these out for me......its practice, not homework. I have a test tomorrow.

  • Calculus -

    I would set my equation as

    N = 500(2)^(t/7) , where 7 is the number of hours

    a)when t=10
    N = 500(2)^(10/7) = 1345.9 or 1346 bacteria

    b) set 3000 = 500(2)^(t/7)
    6 = 2^(t/7)
    t/7 = log6/log2
    t = 7log6/log2 = 18.1

    makes sense ....
    after 7 hours 1000
    after 14 hours 2000
    after 21 hours 4000 , 3000 seems reasonable after 18

  • Calculus -

    sorry I forgot to mention I have to use the equation: p(t)= ce^kt

  • Calculus -

    ok, then change it to

    N = 500(e)^(kt)
    when t = 7
    1000 = 500(e)^(7k)
    2 = (e)^(7k)
    7k = ln2
    k = ln2/7 = .09902

    N = 500(e)^(.09902t)

    now follow the same steps as before
    a) let t=10 .... ( I got exactly the same answer as before)

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question