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    im stuck on this question, can someone help me?
    The rate of increase of a population (p) of micro-organism at time (t) is given by dp/dt = kp
    where k is a positive constant. Given that at t=0 the population was of size 8, and at t=1 the population is 56, find the size of the population at time t=2.

    what i have done so far

    1/kp dp=1 dt

    1/k. Ln(kp)= t+c

    when t=0 p=8

    1/k. Ln(8k)=0+c

    when t=1 p=56

    1/k. Ln(56k) -1 =c
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    If you know the initial population and that the population at t=1 is 56, you can use this to find the constant and formulate an exponential equation in terms of P and t. Then, just substitute t=2 into your equation to find P
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    I'd probably be dividing just by p at the start? 1/p dp = kdt?
    This gives c = ln(8) and works out nicer?

    NB there is a 95% chance this is wrong as I am awful at C4...
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    (Original post by M_17)
    I'd probably be dividing just by p at the start? 1/p dp = kdt?
    This gives c = ln(8) and works out nicer?

    NB there is a 95% chance this is wrong as I am awful at C4...
    i found k as Ln 7 but cant seem to find c.
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    can i cancel both the Ln 7 ?

    1/Ln7 .Ln(8.Ln7)=c
 
 
 
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