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Math Help - Exponential Growth and Decay

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    Exponential Growth and Decay

    Can someone please help me with this? I can't seem to get the right answer. Thanks!

    A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 h there are 10000 bacteria. at the end of 5 h there are 40000 bacteria. How many bacteria were present initially?
    Last edited by turtle; February 27th 2007 at 09:41 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by turtle View Post
    Can someone please help me with this? I can't seem to get the right answer. Thanks!

    Find the amount of time required for a $2000 investment to double if the annual interest rate r is compounded a) monthly, and b) quarterly


    What is r?

    Anyway, i'll give you the general solutions and you can plug in r later.

    In general, if interest is compounded n times per year, we use

    A(t) = A0*(1 + r/n)^nt
    where A(t) is the amount of money at any given time, A0 is the initial amount of money, r is the rate of interest, t is time, n is the number of times per year the money is compounded. Here goes.

    We want A(t) = 2*A0 (since they said they wanted it to double), so we want

    2*A0 = A0*(1+ r/n)^nt
    => 2 = (1 + r/n)^nt .........................divided both sides by A0
    => ln2 = ln(1 + r/n)^nt
    => ln2 = nt*ln(1 + r/n)
    => t = ln2/(n*ln(1 + r/n)), where t is the time in years.

    a) Compounded monthly, => n=12
    so t = ln2/(12ln(1 + r/12)) years

    b) Compounded quarterly, => n=4
    so t = ln2/(4ln(1 + r/4)) years

    c) annually, => n=1
    so t = ln2/ln(1 + r) years
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by turtle View Post
    Can someone please help me with this? I can't seem to get the right answer. Thanks!

    A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 h there are 10000 bacteria. at the end of 5 h there are 40000 bacteria. How many bacteria were present initially?

    Ah, you changed the question! I was wondering about that, why did he call it "exponential growth and decay"
    Last edited by Jhevon; February 27th 2007 at 10:08 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by turtle View Post
    Can someone please help me with this? I can't seem to get the right answer. Thanks!

    A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 h there are 10000 bacteria. at the end of 5 h there are 40000 bacteria. How many bacteria were present initially?
    I hope you dont change the question again while i'm typing this one up.



    For exponential decay, we use the formula P(t) = Po*e^rt, where P(t) is the amount after time t, Po is the initial amount, r is the rate of growth, t is the time.

    We know P(3) = 10000, P(5) = 40000. So lets take a ratio of the two.

    P(5)/P(3) = Poe^5r/Poe^3r = 40000/10000
    => e^2r = 4.................since the Po's cancel, and e^3r into e^5r leaves e^2r
    so lne^2r = ln4
    => 2rlne = ln4
    => 2r = ln4
    so r = ln4/2 = 0.693 approx.

    Now we can find Po by plugging in what we know in either the expression for P(3) or P(5).

    Recall: P(t) = Po*e^rt
    => Po = P(t)/e^rt

    Say you wanted to use P(5):

    Po = 40000/e^(5*0.693) = 1250.9 = 1251 bacteria

    Say you wanted to use P(3):

    Po = 10000/e^(3*0.693) = 1250.5 = 1251 bacteria


    I'm thinking the answer in the back of the book says something like 1250, but that discrepancy would be caused by my rounding off ln4/2
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