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Math Help - Exponential growth + decay

  1. #1
    wombat1
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    Exponential growth + decay

    I'm having some trouble with this hwrk problem, and help would be greatly appreciated. Thanks!

    If a substance decomposes at a rate proportional to the amount of substance present, and if the amount decreases from 40gm to 10gm in 2 hours, then the constant of proportionality is:
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  2. #2
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    Quote Originally Posted by wombat1
    I'm having some trouble with this hwrk problem, and help would be greatly appreciated. Thanks!

    If a substance decomposes at a rate proportional to the amount of substance present, and if the amount decreases from 40gm to 10gm in 2 hours, then the constant of proportionality is:
    Hello,

    the process could be best described by an exponential funktion: f(t)=a \cdot b^t

    where a is the value at the beginning
    b (the base) is the constant of proportionality you are looking for
    t is the time.

    Put in the values you know: 10=40 \cdot b^2

    Solve for b: \frac{10}{40}= b^2 and you'll get: b = 0.5.

    Greetings

    EB
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  3. #3
    MHF Contributor
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    Quote Originally Posted by wombat1
    I'm having some trouble with this hwrk problem, and help would be greatly appreciated. Thanks!

    If a substance decomposes at a rate proportional to the amount of substance present, and if the amount decreases from 40gm to 10gm in 2 hours, then the constant of proportionality is:
    Here is one way.
    "Decomposes at a rate proportional to amount of substance at present"
    Let present be when time is zero.
    So,
    amount of substance at present = "P at t=0" = Po
    amount of substance at time t = "P at t=t" = P of t = P(t)
    P(t) = (Po)*e^(kt) -------------exponential function

    Given:
    at time t=2hrs,
    P(t) = P(2) = 10gm
    Po = 40gm at anytime.

    So,
    10 = 40*e^(k*2)
    10/40 = e^(2k)
    1/4 = e^(2k)
    ln(1/4) = ln[e^(2k)] -----(i)
    ln(0.25) = (2k)*ln(e)
    ln(0.25) = 2k
    k = (1/2)*ln(0.25)
    k = -0.693147181 -------------------answer.

    Or, from (i),
    ln(1/4) = ln[e^(2k)] -------(i)
    ln(1) -ln(4) = (2k)*ln(e)
    0 -ln(2^2) = (2k)*1
    -2ln(2) = 2k
    k = -ln(2) ---------------(ii)
    k = -(0.693147181)
    k = -0.693147181
    Same as above, hence, k = -ln(2).

    Then,
    k = -0.693147181 = -ln(2) ----------if you are using the "e".

    If you are not using the "e",
    P(t) = (Po)*e^(kt)
    P(t) = (Po)*[e^(k*t)]
    P(t) = (Po)*[(e^K)^t]

    To not to use "e", let (e^k) = a = constant of proportionality.
    So,
    P(t) = (Po)*[a^t]
    Plug in the known values,
    10 = 40[a^2]
    10/40 = a^2
    1/4 = a^2
    a = 1/2
    a = 0.5 ---------------answer.
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