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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.2010-12-09
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    @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...2010-12-09
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    @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.2010-12-09
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    @Arturo Magidin $n \ln r = \ln 0.5$ implies $r=e^{\frac{\ln\left(0.5\right)}{n}}$ , or am I misinterpreting the question?2010-12-09
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    @WWright: My point is that if you have $r^n = 0.5$ with $n$ integral and $r$ positive, then you simply get $r=(0.5)^{1/n}$ without the need to use logarithms (or even know about them). (Of course, $e^{\ln(0.5)/n} = (e^{\ln(0.5)})^{1/n} = (0.5)^{1/n}$, same answer). When the unknown is the exponent, logarithms *do* naturally suggest themselves and are usually "an excellent way" to deal with it; but when your unknown is the base raised to an integral power, I don't see why one would think to use logarithms. For sure one *can* use logarithms, I just don't see why one would think to, or want to.2010-12-09
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    @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.2010-12-09
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    @Arturo Magidin I see your point. Now that I think about it, the approach I suggested was quite unnatural (even though it was the first that popped into my head for whatever reason)2010-12-09
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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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