Learning Some Logarithms

A colleague asked how many digits of \(\pi\) I can recite from memory. I realised<br>I still remember the 27.3 digits I memorised in my childhood11 Wait, what does<br>it mean to remember 0.3 digits? I know the digit after the last one is in the<br>range 0–4, which is one bit of information, and the full digit would be just<br>over three bits. Hence, 1/3 or 0.3 digits. Okay but why do I know the digit<br>after the last is in the range 0–4? I deliberately picked a cut-off point that<br>didn&rsquo;t force me to round the last digit up, so that if I wanted to expand and<br>learn even more digits later I wouldn&rsquo;t have to re-learn the last rounded<br>digit.. I was slightly embarrassed to admit this, because of how useless it is<br>to know more than, say, four significant figures of \(\pi\).

There are many other useful values to memorise, e.g. some logarithms would be<br>neat.

Base ten logarithms

I have picked up spaced repetition again, so I now have a way to learn some<br>logarithms basically for free, as well as how to use them for mental maths. The<br>values below are good to know and enough to get started (the rest can be roughly<br>interpolated.)

\(\;\;\;\; \log{1} = 0\)

\(\;\;\;\; \log{2} \approx 0.3\)

\(\;\;\;\; \log{3} \approx 0.5\)

\(\;\;\;\; \log{5} \approx 0.7\)

\(\;\;\;\; \log{8} \approx 0.9\)

\(\;\;\;\; \log{10} = 1\)

A few weeks after I memorised that batch, I added a second batch to fill out<br>some of the gaps. These are probably overkill but convenient once the memories<br>are formed.

\(\;\;\;\; (\log{1.25} \approx 0.1)\)

\(\;\;\;\; (\log{1.6} \approx 0.2)\)

\(\;\;\;\; (\log{2.5} \approx 0.4)\)

\(\;\;\;\; (\log{4} \approx 0.6)\)

\(\;\;\;\; (\log{6} \approx 0.8)\)

\(\;\;\;\; (\log{7} \approx 0.85)\)

\(\;\;\;\; (\log{9} \approx 0.95)\)

Base ten logarithms are particularly convenient because we write numbers in base<br>ten, so we only need to know the logarithms of 1–10 to compute the logarithm of<br>any other number.

For larger numbers, we can approximate using the rule that turns multiplication<br>into addition22 This rule was in fact why Napier invented logarithms in the<br>first place! Multiplication problems are hairy but addition is easier.

\[\log{50} = \log{(5 \times 10)} = \log{5} + \log{10} \approx 0.7 + 1 = 1.7\]

(Incorrect by 0.06 %.)

We can always just round to the nearest number with a bunch of zeroes on it:

\[\log{87234} \approx \log{(8.7 \times 10^4)} = \log{8.7} + \log{10^4} \approx 0.93 + 4 = 4.93\]

(Incorrect by 0.2 %.)

To approximate log 8.7, we see that it is about a third of the way between 8 and<br>10, and since all continuous functions are linear when you look up close, we<br>just go a third of the way from log 8 to log 10, both of which we know!

Smaller numbers work on the same principle:

\[\log{0.055} = \log{(5.5 \times 10^{-2})} = \log{5.5} + \log{10^{-2}} \approx 0.73 - 2 = -1.27\]

(Incorrect by 0.8 %.)

Multiplication

If we need to multiply two numbers, say 365×48, we can do that by instead<br>estimating the equivalent

\[10^{\log 365 + \log 48}\]

where the relevant logs turn it into something like

\[10^{2.55 + 1.68} = 10^{4.23} \approx 17000\]

(Incorrect by 3 %.)

Natural logarithms

If we add another two numbers to our memory we can do even more useful things:

\(\;\;\;\; \log{e} \approx 0.43\)

\(\;\;\;\; \frac{1}{\log{e}} \approx 2.3\)

Now we can change from base-ten logarithms to natural logarithms by plain<br>multiplication with 2.3. Quick example:

\[\ln{57} \approx 2.3 \times \log{57} \approx 2.3 \times 1.77 \approx 4.1\]

(Incorrect by 1 %.)

One neat property of the natural logarithm is that for small \(x\), we can<br>estimate \(\ln{(1 + x)} = x\). So \(\ln{1.03} \approx 0.03\).33 This sort of<br>conversion is convenient when discussing growth rates of lognormal random walks<br>and log-odds differences of small effects.

Powers

Raising a number to the power of another comes with another handy rule, namely<br>that

\[a^b = 10^{b \log{a}}\]

Since we have \(\log{e}\) from before we can estimate

\[e^3 = 10^{3 \log{e}} \approx 10^{3 \times 0.43} \approx 10^{1.3} \approx 20\]

(Incorrect by 0.4 %.)

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Square roots

Square roots are, of course, just powers under the hood.

\[\sqrt{512} = 512^{0.5} = 10^{0.5 \times \log{512}} \approx 10^{0.5 × 2.7} \approx 10^{1.3} = 20\]

(Incorrect by 13 % – the main imprecision enters the picture in the sloppy<br>division of 2.7 by two. If we compute with 1.35 instead of 1.3, we get a quarter<br>of the way between 20 and 30, i.e. 23, which is inaccurate only by 2 %.)

What&rsquo;s neat is you can use the same technique for third roots, which are<br>otherwise much trickier to guess!

Next steps

I&rsquo;m not fast at doing this in my head yet, but I can at least do it without a<br>calculator, which is more than I was able to...