Effective half-life multiple exponential decay

  • Thread starter NanakiXIII
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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
Jmf
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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:

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