How do I give a good estimate for $3^{3.8}$ or $3^{308}$ or $3^{3.7}$ etc. quickly and without a calculator? How do I give a good estimate for $3^{3.8}$ or $3^{3.08}$ or $3^{3.7}$ etc. quickly and without a calculator?
 A: Back of the envelope estimate of $3^{3.8}$ to follow:
We know $\log 3$ is slightly bigger than $1$.
Let $f(x) = 3^x$, so that $f'(x) = 3^x \log 3$.  
If $x = 4$ and $dx = -.2$ then $y = 81$ and $dy = 3^4 \log 3 (-.2) = -16.2 (\log 3)$. 
So a not-too-accurate guess using differentials could be $81 - 16.2(\log 3)$. Since $\log 3$ is slightly bigger than $1$ let's just call it $81 - 17 = 64$.
If you go a bit further and use the fact that $y = 3^x$ is convex, we know the initial estimate is going to be too low, so perhaps bump it up a bit to 65.
A: The binomial theorem shows that $(1+0.2)^5<3<(1+0.3)^5$, and the difference is about equally large on either side of the inequality. So let's say $\sqrt[5]3\approx1.25$. Then
$$
\frac{81}{3^{0.2}}\approx 81\cdot\frac 45=64.8
$$
A: The following approach is "purely rational", i.e., does not require the floating point computation of logarithms, etc.
Assume that we want to compute $3^x$ for arbitrary real $x>0$. Note that
$$3^x=\left({1\over 3}\right)^{-x}=\left(1-{2\over3}\right)^{-x}=\sum_{k=0}^\infty{-x\choose k}\left(-{2\over3}\right)^k=1+{2x\over3}+{2x(x+1)\over 9}+\ldots\quad.$$
The series is convergent for all $x>0$ (even for all $x\in{\mathbb C}$). When $x>0$ its terms start to decrease when $k>2x-3$.
A: Do you know how these things are calculated in a calculator, they use the taylor approximation. 
https://en.wikipedia.org/wiki/Taylor_series
So if you could memorize up the first few terms of the series of $a^x$
$a^x = 1 + x log(a) +\frac{1}{2} x^2 log^2(a) + \frac{1}{6}x^3 log^3(a)....$
https://www.wolframalpha.com/input/?i=expand+a%5Ex%5C
and if you know the $log_{10}$ values, this can be used.
