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I feel like this should have a rather simple solution using logarithms, but I don't quite know where to start.

Basically, every day I am going to multiply a given quantity by $r$, where $r < 1$.

Obviously, if $r$ is 0.5, then my half-life $n$ is 1 day. (This is also stated to remove any ambiguities on what counts as a day/range)

What is the relationship between $r$ and $n$?

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Using logarithms is an excellent way to approach this problems. You know that $r^{n} =0.5$. So if you remember the basic properties of logarithms, this should get you going in the right direction. – WWright Dec 9 '10 at 3:47
@WWright: But since the unknown is $r$, not $n$, why would one want to use logarithms? Logarithms suggest themselves when your unknown is in the exponent, or perhaps when the exponent is not integral, as it is here... – Arturo Magidin Dec 9 '10 at 3:55
@Justin L.: Incidentally, I don't know how you learned it, but a nice way to view an exponential decay model with half-life $T$ is as $f(t)=a2^{-t/T}$. This way it is easily seen that $f(t+T)=a2^{-t/T-1}=\frac{1}{2}a2^{t/T}=\frac{1}{2}f(t)$, and looking at $f(1)$ with $T=n$ is one way to answer your question. – Jonas Meyer Dec 9 '10 at 4:00
@Arturo Magidin $n \ln r = \ln 0.5$ implies $r=e^{\frac{\ln\left(0.5\right)}{n}}$ , or am I misinterpreting the question? – WWright Dec 9 '10 at 5:40
@WWright: What you've written is correct and agrees with what Justin got to (see the comments on my answer), but the answer is not in a particularly nice form, largely due to applying logs when the variable to be solved for wasn't in the exponent. – Isaac Dec 9 '10 at 5:43

Suppose $x$ is the quantity, and you want it to decay to $\frac{1}{2}$ after $n$ discrete steps (whether they be minutes, hours, days, years, eons, or nanoseconds), and you want to model that by multiplication by $r$.

You start with $x$. After one time interval, you want to have $rx$. After two time intervals, you want $r(rx) = r^2x$. After three time intervals, $r(r^2x)=r^3x$. After $n$ times intervals, $r^nx$. Since you want $r^nx = \frac{1}{2}x$, you can now solve for $r$.

This is discrete decay, though, not continuous decay. It is exactly analogous to the difference between continuous interest compounding, and interest compounding in discrete time intervals. "Half-life" is a term usually reserved for continuous decay, so you'll want to be careful with that.

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Suppose you start with a quantity of 1 (whatever that means). You know that after $n$ days, there should be $\frac{1}{2}$ left. This should give you an equation that you can solve for $r$ (in terms of $n$ or with a specific value of $n$).

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I appreciate that you leave the rest up to me so that I don't feel any stupider than I already feel, heh. – Justin L. Dec 9 '10 at 3:48
For a long time, I've felt that the way exponential growth/decay and half-life/doubling-time are taught don't quite mesh for a lot of people. I was hoping to give you enough to get you going, figuring that you know you can drop a comment here if what I said isn't getting you to the result. – Isaac Dec 9 '10 at 3:53
I seem to end up with $r = e^{\frac{ln (0.5)}{n}}$...which I guess is right, but doesn't seem to be as pretty as I feel it should. I really only have to compute it once, before my program starts iterating, but is there a prettier solution? – Justin L. Dec 9 '10 at 4:02
@Justin: $r = e^{\frac{ln (0.5)}{n}}=\left(e^{\ln(0.5)}\right)^{\frac{1}{n}}=0.5^{\frac{1}{n}}$ is probably what you want. $r^n=0.5\implies r=0.5^{\frac{1}{n}}$. – Isaac Dec 9 '10 at 4:05

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