Approximation of $e^{-x}$ Is there a method to mentally evaluate $e^{-x}$ for $x>0$? Just to have an idea when computing probabilities or anything that is an exponential function of some parameters.
 A: If $x$ is "small enough", I like using the $(2,2)$ Padé approximant, which is easily cast into a memorable form:
$$\exp\,x\approx \frac{(x+3)^2+3}{(x-3)^2+3}$$
For $|x| < \frac12$, the absolute difference between the approximant and the true function is $< 8\times 10^{-5}$. Pretty good approximation for a mere rational function... of course, as with all such approximants, the further you go from $0$, the less accurate it becomes.
A: As a supplement to Henry's answer, consider these rational approximations to $\log_{10}(e)$ and $\log_{2}(e)$.
Two of the approximants for the continued fraction for $\log_{10}(e)$ are $\frac{3}{7}$ (low and not as good as $.43$) and $\frac{10}{23}$ (high but better than $.43$). So if it makes computation easier, you can try
$$
10^{-10x/23}<e^{-x}<10^{-3x/7}\tag{1}
$$
when $x>0$. The order in $(1)$ is reversed for $x<0$.
Two of the approximants for the continued fraction for $\log_{2}(e)$ are $\frac{10}{7}$ (low and not as good as $1.44$) and $\frac{13}{9}$ (high but better than $1.44$). So if it makes computation easier, you can try
$$
2^{-13x/9}<e^{-x}<2^{-10x/7}\tag{2}
$$
when $x>0$. The order in $(2)$ is reversed for $x<0$.
A: Dilip was asking in the comments about "Feynman's method"; since this is too long for a comment, I shall be quoting the relevant paragraphs here:

Lucky Numbers
One day at Princeton I was sitting in the lounge and overheard some mathematicians talking about the series for $e^x$, which is
  $1+x+x^2/2!+x^3/3!$ Each term you get by multiplying the preceding
  term by $x$ and dividing by the next number. For example, to get the
  next term after $x^4/4!$ you multiply that term by $x$ and divide by
  $5$. It's very simple.
When I was a kid I was excited by series, and had played with this
  thing. I had computed $e$ using that series, and had seen how quickly
  the new terms became very small.
I mumbled something about how it was easy to calculate $e$ to any
  power using that series (you just substitute that power for $x$).
"Oh yeah?" they said. "Well, then what's $e$ to the $3.3$?" said some
  joker—I think it was Tukey.
I say, "That's easy. It's $27.11$."
Tukey knows it isn't so easy to compute all that in your head. "Hey!
  How'd you do that?"
Another guy says, "You know Feynman, he's just faking it. It's not
  really right."
They go to get a table, and while they're doing that, I put on a few
  more figures. "$27.1126$," I say.
They find it in the table. "It's right! But how'd you do it!"
"I just summed the series."
"Nobody can sum the series that fast. You must just happen to know
  that one. How about $e$ to the $3$?"
"Look," I say. "It's hard work! Only one a day!"
"Hah! It's a fake!" they say, happily.
"All right," I say, "It's $20.085$."
They look in the book as I put a few more figures on. They're all
  excited now, because I got another one right.
Here are these great mathematicians of the day, puzzled at how I can
  compute $e$ to any power! One of them says, "He just can't be
  substituting and summing—it's too hard. There's some trick. You
  couldn't do just any old number like $e$ to the $1.4$."
I say, "It's hard work, but for you, OK. It's $4.05$."
As they're looking it up, I put on a few more digits and say, "And
  that's the last one for the day!" and walk out.
What happened was this: I happened to know three numbers—the logarithm
  of $10$ to the base $e$ (needed to convert numbers from base $10$ to
  base $e$), which is $2.3026$ (so I knew that $e$ to the $2.3$ is very
  close to $10$), and because of radioactivity (mean-life and
  half-life), I knew the $\log$ of $2$ to the base $e$, which is
  $.69315$ (so I also knew that $e$ to the $.7$ is nearly equal to $2$).
  I also knew $e$ (to the $1$), which is $2.71828$.
The first number they gave me was $e$ to the $3.3$, which is $e$ to
  the $2.3$—ten—times $e$, or $27.18$. While they were sweating about
  how I was doing it, I was correcting for the extra $.0026$—$2.3026$ is
  a little high.
I knew I couldn't do another one; that was sheer luck. But then the
  guy said $e$ to the $3$: that's $e$ to the $2.3$ times $e$ to the
  $.7$, or ten times two. So I knew that it was $20.$something, and
  while they were worrying about how I did it, I adjusted for the
  $.693$.
Now I was sure I couldn't do another one, because the last one was
  again by sheer luck. But the guy said $e$ to the $1.4$, which is $e$
  to the $.7$ times itself. So all I had to do is fix up $4$ a little
  bit!
They never did figure out how I did it.
...

There's less to it than meets the eye. ;)
A: $e^{-x} \approx  2^{-1.44x} \approx 10^{-0.43x}$
where $\log_2(e) \approx 1.44$ and $\log_{10}(e) \approx 0.43$.   
A: A diagram may also help...
A: For $e^x$ and limiting to fractions that are low multiples of $\frac{1}{3}$ ... one can approximate fairly well by knowing $3^n$ fairly well,
$$2.714 \approx 3^3/10$$
Using this you can then obtain results via,
$$e^n \approx 2.714^n \approx \left(\frac{3^{3}}{10}\right)^n = \frac{3^{3n}}{10^n}$$
Clearly $10^n$ isn't going to be an issue. Most mathematicians will probabaly be familiar up to $3^4=81$ by memory and calculating up to $3^6$ isn't too much trouble.
Examples:
$$
e^2 \approx 729 / 100 \approx 7.3\\
e^2 = 7.4
$$
$$
e^{3} = e^2e^1 \approx 7.4 \times 2.7 \approx 3/4\times 27 \approx 81/4 \approx 20\\
e^3 = 20.09
$$
you could also get to near 20 by knowing that $3^6$ is 19000 and something
Finding the negative powers just flip the relation.
A: I use $e^{-x} \approx 1-x$ for $x\leq 1$ and $e^{-x}\approx 0$ for $x\geq 1$ and it is truly powerful.
