# Powers of 2 and logs base 10

A trick I learned from Colin Wright:

Take the first nine powers of 2 and place them in lexical order; then put a decimal point after the first digit.

OK, Colin, here you go:

- 1.28
- 1.6
- 2.
- 2.56
- 3.2
- 4.
- 5.12
- 6.4
- 8.

These numbers are approximately $10^{0.1}$, $10^{0.2}$, etc., up to $10^{0.9}$.

Huh! Isn’t that nice? Why does *that* work? You might want to have a think about it. I’ll spoiler it below the line.

Well, it boils down to the handy numerical coincidence that $2^{10} \approx 10^3$.

That means that $2 \approx 10^{0.3}$, $4 \approx 10^{0.6}$ and $8 \approx 10^{0.9}$, before we get into anything complicated.

Then we know that $16 \approx 10^{1.2}$ – and if we divide both by 10, we get $1.6 \approx 10^{0.2}$. Continuing in this vein, we get $3.2 \approx 10^{0.5}$ and $6.4 \approx 10^{0.8}$.

Then we get $12.8 \approx 10^{1.1}$, so $1.28 \approx 10^{0.1}$, leading to $2.56 \approx 10^{0.4}$ and $5.12 \approx 10^{0.7}$ – and that’s the lot!

In fact, we can do something very similar with powers of 5 (I struggle to remember these, but they do have their uses).

- 1.25
- 1.5625
- 1.953125
- 2.5
- 3.125
- 3.90625
- 5
- 6.25
- 7.8125

These are *also* good estimates for $10^{0.1}$, etc., for very similar reasons – if $2^{10} \approx 10^3$, then $5^{10} \approx 10^7$.

I mean, that’s a pretty solid fit.

In fact:

- the powers of 2 are slight overestimates and the powers of 5 are slight underestimates.
- the smaller the (original) power involved, the better the estimate – 5 is a better guess for $10^{0.7}$ than 5.12 is, and 4 is a better approximation for $10^{0.6}$ than 3.90625.
- you might well know that $10^{0.5} \approx 3.16$ – that’s roughly the arithmetic mean of 3.125 and 3.2 ((and it’s exactly the geometric mean.))

So there you go! A nice little trick and the reason behind it.