What is the method for mentally computing $3^{3.5}$ and similar calculations? What is the method for mentally computing an approximation of $3^{3.5}$ and similar calculations?
(without using any calculator)
The best I did is:
$3^{3.6}=e^{3.6ln(3)}=e^{3.6*1.098}  \approx e^{3.6*1.1}=e^{3.96} \approx e^4=54.5$
 A: Methods for mental calculation  often depend on knowing facts.  For this problem, knowing $\sqrt 3\approx 1.732$ makes it easy.  $$3^{3.5}\approx 3^3\sqrt 3=27\cdot 1.732=46.764$$
where I did the final multiply mentally and I should delete the last digit because I only have four digits accuracy in the input.  It is also important to be extremely facile with $(1+x)^n \approx 1+nx$ for $nx \ll 1$.  This lets you adapt things to the facts you know. 
A: Memorize three logarithms
Here’s how I do it. I’ve got the base-$10$ logarithms for the integers $2$, $3$, and $7$ memorized:
$$\begin{array}{|r|r|} \hline
n & \log n \\ \hline
2 & .301 \\ \hline
3 & .477 \\ \hline
7 & .845 \\ \hline
\end{array}$$
That’s all you need to get the logarithm for the first digit of any number, because you can easily deduce the logs for the other first digits from $.301$ and $.477$. For example, $\log 6 = \log 2 + \log 3 \approx 0.301 + 0.477 = 0.778$.
So, I can use logarithms to reduce exponentiation to multiplication, which I can do quickly in my head if I only consider the first couple digits. First I remember:
$$\log 3 \approx 0.477$$
$0.477$ is a little less than half, and half of $3.5$ is $1.75$, so here’s another quick mental approximation:
$$3.5 \times 0.477 \approx 1.7$$
Now I just need to figure out the antilogarithm of $1.7$. The characteristic (the part to left of the decimal) is $1$, so the antilog is in the $10$’s, i.e. in the range $[10, 100)$. The mantissa is $0.7$, which I recognize as just about $0.301$ less than $1$. Referring to the memorized table again, $0.301 \approx \log 2$, so $10^{0.7} \approx 10 \div 2 = 5$. That gives me the first digit of the answer. So, I conclude:
$$3^{3.5} \approx 50$$
A calculator reveals that the true answer is $46.765372$. So, not bad for a few seconds of mental fiddling.
The trade-off
Notice that Ross Millikan’s approximation is much more accurate. His approach relies on having a large library of exploitable facts memorized, and capitalizing on the most relevant ones. That’s also how lightning calculators do it. The method in this answer is crude but very easy to learn. You just memorize three numbers and you’re set for life. It was told to me by a chemistry professor long ago. It’s good for quickly estimating a great many calculations in physics, chemistry, and the like that involve exponents or roots. Sometimes it can be hard to estimate the antilogarithm. I suppose I could memorize that $10^{0.1} \approx 1.26$ and maybe a couple more, but I never have.
A: Based on the facts you had already memorized,
\begin{multline}
3^{3.5} = e^{3.5\ln3} \approx e^{3.5(1.098)} = e^{3.5(1.1 - 0.002)} \\
 = e^{3.85 - 0.007} = e^{3.843} = e^4 e^{-0.157}
 \approx 54.5 \times 0.843 = 45.9435 \approx 46,
\end{multline}
where the factor $0.843$ comes from the Taylor expansion of $e^x$ up to the term in $x$ (discarding $x^2$ and later terms), so it is an underestimate.
Guessing that we have at most two significant terms, we round up.
(You may also take some shortcuts with $54.5 \times 0.843,$ which is hard to do mentally; but try to round upward.)
If we also take the $x^2$ term of the Taylor expansion, we get 
$$e^{-0.157} \approx 1 - 0.157 + \frac{0.157^2}2 
\approx 1 - 0.157 + \frac{0.025}2 \approx 0.855,$$
and the final approximation is 
$3^{3.5}\approx 54.5\times 0.855 = 46.6$
(or whatever you use to approximate $54.5\times 0.855$).
That's still not quite as good (or even as easy to do) as Ross Millikan's method, but it generalizes well.
