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Math Help - Exponential Growth and Decay

  1. #1
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    Exponential Growth and Decay

    I'm having some trouble with the following problems:

    1) Radioactive decay problem: What percent of a present amount of radioactive radium (226Ra) will remain after 900 years? (Half-life: 1599 years)

    2) Radioactive decay problem: Find the half-life of a radioactive material if after 1 year 99.57% of the initial amount remains.

    3) In 1990, the population of Mexico was 89 million and growing at the rate of 1.8% per year. In 1990, the US population was 250 million and growing at the rate of .7% per year. If these rates continue, when will the 2 nations have the same population?

    Please help if you can, thanks alot.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cupcakelova87 View Post
    I'm having some trouble with the following problems:
    For radioactive decay we use the formula:

    A(t) = A0*e^(-rt)

    where A(t) is the amount left after time t, A0 is the intitial amount, r is the rate of decrease, t is the time

    1) Radioactive decay problem: What percent of a present amount of radioactive radium (226Ra) will remain after 900 years? (Half-life: 1599 years)
    r = ln2/t, where t is the half-life
    => r = ln2/1599 = 0.000433487

    we want A(t) when t is 900, that is, we want A(900)

    A(900) = A0 *e^(-0.000433487(900)) = A0(0.676963)

    so A(900) = (0.676963)A0

    so A(900) is 67.69% of A0. this is the percent remaining




    2) Radioactive decay problem: Find the half-life of a radioactive material if after 1 year 99.57% of the initial amount remains.
    assume we start with 1 unit of the radioactive substance, then after 1 year, 0.9957 units are left.

    using A(t) = A0*e^(-rt)

    we have A(1) = 0.9957 = 1*e(-r(1))
    => e^-r = 0.9957
    => ln(e^-r) = ln(0.9957)
    => -r = ln(0.9957)
    => r = - ln(0.9957)
    => r = 0.004309271

    now t = ln2/r, where t is the half life
    => t = ln2/0.004309271
    => t = 160.8502 years
    Last edited by Jhevon; April 8th 2007 at 08:18 PM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cupcakelova87 View Post

    3) In 1990, the population of Mexico was 89 million and growing at the rate of 1.8% per year. In 1990, the US population was 250 million and growing at the rate of .7% per year. If these rates continue, when will the 2 nations have the same population?
    For exponential growth, we use the same formula for decay, but with one difference, the power of e is positive. that is, we use the formula:

    P(t) = P0*e^(rt)

    where P(t) is the population at time t, P0 is the initial population, r is the rate of growth, t is the time

    now for mexico, we are given r = 0.018 and the initial population in 1990 is P0 = 89 million.
    that is, for Mexico, P(t) = 89000000e^0.018t

    now for the US, we are given r = 0.007 and the initial population in 1990 is 250 million
    that is, for the US, P(t) = 250000000e^0.007t

    when will they have the same population?
    when 89000000e^0.018t = 250000000e^0.007t of course
    now all we do is solve for t
    => e^0.018t = (250/89)e^0.007t ..........i divided both sides by 89 million
    => e^0.018t/e^0.007t = 250/89 ...........i divided both sides by e^0.007t
    => e^0.011t = 250/89 ........................when dividing the same base we subtract the powers. 0.018t - 0.007t = 0.011t

    now we have an exponential equation. how do we solve this? we log both sides of course

    => lne^0.011t = ln(250/89)
    => 0.011t = ln(250/89)
    => t = ln(250/89)/0.011
    => t = 93.89 ~= 94 years

    so the populations will be equal after about 94 years, that is in the year 2084
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  4. #4
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    Thanks Jhevon! I have one more problem...

    On the Richter scale, the magnitude R of an earthquake of intensity I is given by R=(ln I - ln Io)/ln 10
    where Io is the minimum intensity used for comparison. Assume Io=1.
    a) Find the intensity of the 1906 SF earthquake in which R=8.3.
    b) Find the factor by which the intensity is increased when the value of R is doubled.
    c) Find dR/dI.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Sorry, I never noticed this question 'till just now

    Quote Originally Posted by cupcakelova87 View Post
    Thanks Jhevon! I have one more problem...

    On the Richter scale, the magnitude R of an earthquake of intensity I is given by R=(ln I - ln Io)/ln 10
    where Io is the minimum intensity used for comparison. Assume Io=1.
    a) Find the intensity of the 1906 SF earthquake in which R=8.3.
    If Io = 1, then ln(Io) = 0
    So we have R = ln(I)/ln(10)

    => ln(I) = ln(10)R
    => I = e^(R*ln(10)) = e^ln(10^R) = 10^R
    When R = 8.3
    I = 10^8.3 ~= 1.995 x 10^8

    b) Find the factor by which the intensity is increased when the value of R is doubled.
    when R = 2*8.3 = 16.6
    I = 10^16.6 ~= 3.981 x 10^16

    so I increased by a factor of (3.981 x 10^16)/(1.995 x 10^8) ~= 1.995 x 10^8 ~= 2 x 10^8

    so basically, I is squared when R is doubled


    c) Find dR/dI.
    R = ln(I)/ln(10)
    => dR/dI = (1/ln(10))*(1/I) = 1/ln(10)*I
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