Euler's constant, $e$ (about 2.718 281 828) is one of the most important numbers in maths -- both pure and applied. (Thinking about my final year university courses, the only one I'm pretty sure had no use for $e$ was History of Maths, and frankly that was an oversight.)

As a budding mathematical ninja, you're doubtless keen to learn how to estimate powers of $e$ for pleasure and profit. Unfortunately, only pleasure is likely to come of it, unless you become a maths tutor.

Now: it's easy enough to estimate $e$ -- it has a nice, memorable decimal expansion, and depending on how roughly you want to play, you can look at it as 3 (-10%), 2.7(+0.7%) or $\frac{30}{11}$ (-0.3%).

However, your estimates for powers of $e$ are likely to be off by more than you're used to in ninja maths -- this is one situation where small errors add up fast. If you're about as good as me, you'll be happy to get things within about 5%.

If you want to know $e^3$, you might pick the first one, and say 'it's 27 less 30% -- take away about 8, so 19. It's actually 20.08 -- not brilliant, but ok for a ballpark figure. It's not really obvious how to do $2.7^3$ -- unless you know that $27^3$ is about 19,700, so $2.7^3$, plus 2.1%, would be 19.7 plus about 0.4, or 20.1. That's bang on.

Alternatively, you can do $\frac{30^3}{11^3}$, which gives $\frac{27,000}{1,331}$. Multiplying top and bottom by 3 gives $\frac{81,000}{3,993}$; that's a shade over $\frac{81}{4}$, or 20.25 -- less about 1% to make up for the original estimate, making 20ish. The only limit is your number handling!

If you're hot on your natural logs, you can reverse-engineer powers of $e$ from there, too -- if you want to know $e^{3.5}$, you can ask '$\ln$ of what is 3.5? Well, that's $0.7 \times 5$, so it must be about $2^5$ or 32. (It's 33.11 -- not bad).

When you're working with negative powers of $e$, the fractional version ($e \approx \frac{30}{11}$) comes into its own - because $e^{-1}$ is just $\frac{11}{30}$, and it's easy to take powers of both of those.

For instance, $e^{-2}$ is just $\frac{11^2}{30^2}$, or $\frac{121}{900}$. That's a bit more than $\frac{40}{300}$, or 0.133. (In fact, it's 0.135). And $e^{-1}$ itself comes up a lot: $\frac{11}{30}$ is 0.367, while $e^{-1}$ is 0.368. I could live with that as an estimate!

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

## Cav

As if you missed this pun opportunity : “… this is one situation where small errors add up fast”(one could say exponentially)….

## Colin

*Puts 50p in the missed-pun-pot*

## Rob C.

You can do a mental, order-of-magnitude estimate if you know properties of natural logs and that ln (10) = 2.3. Consider e^x = 10^y. Therefore, ln (e^x) = ln (10^y). Simplifying, x = ln (10^y). Next, x = y ln 10, or y = x/ln 10. Since ln10=2.3, y=x/2.3. So substituting that back into the original equation, e^x=10^(x/2.3). So if you can estimate x/2.3 to within one whole number, you can get an order-of-magnitude estimate for e^x.

## Colin

Good shout! Thanks, Rob đź™‚