"The Mathematical Ninja is currently on sabbatical. Leave a message after the tone... or else!"

Oh dear! How are we going to figure out $e^e$ now? Let alone $e^{-\frac{1}{e}}$? We'll just have to roll up our sleeves and get our thinking hats on, that's all.

### OK, $e^e$

First of all, we know that $e$ is about 2.72. We're not going any more accurately than that here, because that would be ridiculous.

We can put some bounds on our answer: $3^3$ is 27, so $e^e$ is clearly less than that. Similarly, "more than 4" is true, if not necessarily useful.

How about any logarithms we know? $\ln(16) = 4\ln(2) \approx 4 \times 0.7$ (less 1%), which is $2.8 - 0.028$, or 2.772 - that gives us a new upper bound.

Similarly, $\ln(15) = \ln(3) + \ln(5)$. I don't know those so closely, but that's around 1.10 + 1.61 or 2.71.

"Somewhere between 15 and 16" is pretty good, I'd say -- but can we get closer?

We know that $e^{2.71} \approx 15$, and if we multiply that by $e^{0.01} \approx 1.01$, that gives us 15.15.

Similarly, $e^{2.77} \approx 16$, and multiplying by $e^{-0.05} \approx 0.95$ leads us to 15.2.

Somewhere in the region of 15.15 or 15.2 seems reasonable. And if the Ninja is away, we can reach for the... ohoho, booby-trapped calculator! Nice try, sensei!

15.154.

### What about $e^{-\frac{1}{e}}$?

This came up when trying to find the turning point of $y=x^x$, in case you're interested.

We know what $\frac{1}{e}$ is, it's about 0.368. What has a logarithm of about 0.368?

Well, $\ln(2) \approx 0.693$, so $\ln\br{\sqrt{2}} \approx 0.347$, which is in the right ballpark: $e^{0.347} \approx \sqrt{2}$. If we add 2% to that, we get 1.414 + 0.028, or 1.442, making $e^{0.367}$, which is close enough for me.

However, we want the reciprocal of that. I'd probably call 1.442 "roughly $\frac{13}{9}$", so $e^{-\frac{1}{e}} \approx \frac{9}{13} \approx 0.692$. With all that approximation, I'd have taken anything close to 0.7, but the calculator gives me 0.6922 and a nasty electric shock. Thanks, sensei!

* Edited 2019-09-23 to fix a LaTeX error and to recategorise. Thanks, Adam!

## 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.