# Effective half-life multiple exponential decay

1. Mar 14, 2010

### NanakiXIII

Say we have a decay of the form

$$A e^{-a x} + B e^{-b x}$$.

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

$$A e^{-a x} + B e^{-b x} = \frac{A+B}{2}$$.

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 $$t$$, then a decay given by

$$A e^{-a c x} + B e^{-b c x}$$

has half-life $$\frac{t}{c}$$. 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. Mar 15, 2010

### Jmf

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

$$f(x) = A e^{-a x} + B e^{-b x}$$.

and let

$$g(x) = A e^{-a c x} + B e^{-b c x}$$.

then:

$$g(x) = f(cx)$$

or equivalently:

$$g(\frac{y}{c}) = f(y)$$

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

$$f(t) = \frac{f(0)}{2}$$

then:

$$f(t) = g(\frac{t}{c})$$

Last edited: Mar 15, 2010