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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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    @WWright: The reason, very likely, is the knee-jerk reflex that "half-life$\Rightarrow$exponentials$\rightarrow$logarithms". We all get some of it after taking (or teaching) Calculus.2010-12-09

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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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    @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. $r^n=0.5\implies r=0.5^{\frac{1}{n}}$.2010-12-09