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Effective half-life multiple exponential decay

  1. Mar 14, 2010 #1
    Say we have a decay of the form

    [tex]A e^{-a x} + B e^{-b x}[/tex].

    I haven't had much luck trying to calculate the half-life of such a decay (I'm not sure it's possible, analytically), i.e. solve

    [tex]A e^{-a x} + B e^{-b x} = \frac{A+B}{2}[/tex].

    However, if that's not possible, I'm wondering whether there might still be a way to prove that if the above decay has half-life [tex]t[/tex], then a decay given by

    [tex]A e^{-a c x} + B e^{-b c x}[/tex]

    has half-life [tex]\frac{t}{c}[/tex]. This seems to be true empirically and would make sense, I think. Does anyone have an idea how one might prove it to be true, though?
     
  2. jcsd
  3. Mar 15, 2010 #2

    Jmf

    User Avatar

    Not really a 'proof', but I think it's rather obvious:

    If f(t) is a function, then f(ct) is that function scaled to 1/c of it's size in the t-direction.

    So if we plot the graph of these two functions, the point at which f(ct)=f(0)/2 will clearly be at 1/c times the point where f(t) = f(0)/2.



    EDIT: I just read the above back to myself and it sounded a bit patronizing to say this was 'obvious' - sorry.

    By way of a proof I suppose you could say let

    [tex]f(x) = A e^{-a x} + B e^{-b x}[/tex].

    and let

    [tex]g(x) = A e^{-a c x} + B e^{-b c x}[/tex].

    then:

    [tex]g(x) = f(cx)[/tex]

    or equivalently:

    [tex]g(\frac{y}{c}) = f(y)[/tex]

    So then if 't' is our half life of f, i.e. when:

    [tex]f(t) = \frac{f(0)}{2}[/tex]

    then:

    [tex]f(t) = g(\frac{t}{c})[/tex]
     
    Last edited: Mar 15, 2010
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