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I often find myself in need of a quick approximation to a base-2 logarithm of an integer, e.g. $\log_2 3$ or $\log_2 5$. While I can always reach for a calculator (or computer), I'd quite like to be able to derive one quickly using pen and paper, whenever I need it.

Ideally, such an approximation would be in the form of a sequence of rational numbers that approaches the target value, and which is easy to calculate by hand. Does anyone know of such a sequence?

To give an idea of what I mean: if I ever find myself needing to approximate the Golden ratio, I can simply write down the Fibonacci sequence, pick two consecutive terms and take the ratio between them. This gives successively better rational approximations to $\phi$ and I can easily calculate it even without a pen, since it requires only addition. While I imagine there isn't quite such an easy way to approximate $\log_2 3$ (or $\log_2 5$ or $\log_2 n$), I'm looking for something as close to that as possible.

(If such a thing exists only for natural logarithms it would still be helpful, but base 2 is greatly preferred.)

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  • $\begingroup$ The golden ratio has simple rational approximations because its continued fraction has a nice form. The continued fraction for $\log_2n$ does not have such a nice form (except in the trivial case, where $n$ is a power of 2). $\endgroup$ – Gerry Myerson Oct 16 '17 at 4:49
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It depends on how accurate you want to be. Knowing a few values can help a lot if low accuracy is acceptable. If you know $\ln 2\approx 0.693$ you can convert natural logs and base $2$ logs at will. A big friend is $2^{10} =1024 \approx 10^3$ which is reflected in the fact that $\log_{10}2 \approx 0.30103$ which is very close to $0.3$. Then if you learn $\log_2 (3) \approx 1.585$ and $\log_2 (5) \approx 2.322$ you can get close to any number you want. Once you find a base $2$ log for some number $p$ close to your target $q$ you can use $\log_2 \frac qp =\log_2 q-\log_2 p=\ln 2 \ln(1+\frac {q-p}p)\approx \ln 2 \frac {q-p}p$

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  • $\begingroup$ +1 for some useful tricks. $2^{10}\approx 10^3$ is a good friend of mine already :) $\endgroup$ – Nathaniel Oct 16 '17 at 6:58

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