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Math Help - Exponential growth

  1. #1
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    Exponential growth

    could anyone give me a hand with these please



    1 At the beginning of 1995, an investor decided to invest 6 000 in a plan that grew at an average rate of 6% per year. Show that, if this percentage rate of increase is maintained for 10 years, the value of the original investment will be about 10 745.


    2 A population of bacteria is growing exponentially. Explain briefly what this means, using a sketch graph to illustrate your answer.

    (i) At time t hours, the size, P, of the population of bacteria is given by P = 1000 (2.8)^0.02t. State the value of P when t = 0. Find the value of t when the size of the population is twice its initial value.

    (ii) A second population of bacteria doubles in size every 12 hours. Find the number of hours it takes for this population to increase in size by a factor of 100.
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  2. #2
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    At the beginning of 95,
    Principal,
    p = 6000
    Interest,
    I = \frac{6}{100}
    Final Amount at the end of 95,
     A= p + p(I)
     A= 6000 + 6000(\frac{6}{100})
     A= 6000(1+\frac{6}{100})

    At the beginning of 96,
    Principal,
    p = 6000(1+\frac{6}{100})
    Interest,
    I = \frac{6}{100}
    Final Amount at the end of 96,
     A= p + p(I)
     A= 6000(1+\frac{6}{100}) + 6000(1+\frac{6}{100})(\frac{6}{100})
    A= 6000(1+\frac{6}{100})(1+\frac{6}{100})
    A= 6000(1+\frac{6}{100})^2

    At the beginning of 97,
    Principal,
    p = 6000(1+\frac{6}{100})^2
    Interest,
    I = \frac{6}{100}
    Final Amount at the end of 97,
     A= p + p(I)
     A= 6000(1+\frac{6}{100})^2 + 6000(1+\frac{6}{100})^2(\frac{6}{100})
    A= 6000(1+\frac{6}{100})^2(1+\frac{6}{100})
    A= 6000(1+\frac{6}{100})^3

    So, from here we can safely deduce that after n years, the amount will be
    A= 6000(1+\frac{6}{100})^n

    so after 10 years
    A= 6000(1+\frac{6}{100})^10
    A= 10745

    This however is the long approach to doing things. If you can identify that it is a geometric progression from the question, you can straight use the Tn = ar^n formula

    For the second set of problem
    i)
    P = 1000(2.8)^0.02t
    when t = 0
    p = 1000(2.8)^0.02(0)
    p = 1000

    when p = 2000,
    2000 = 1000(2.8)^0.02t
    2 = 2.8^0.02t
    ln 2 = 0.02t(ln 2.8)
    t = ln 2/0.02(ln2.8)

    ii)
    let the initial population size = X
    Population size = (X)(2^n), where n= number of times the bacteria multiplied

    Given information, population size is p = 100X

    hence,
    100X = X(2^n)
    n ln 2= ln 100
    n = ln 100/ ln 2
    n = 6.644 times

    therefore time taken is 6.644 x 12 = 79.728 hours
    Last edited by bryankek; March 22nd 2010 at 08:55 AM.
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