Feynman's Trick for Approximating $e^x$ Feynman knew how to approximate $e^x$ for small values of $x$ by noting the fact that
\begin{align*} 
& \log10=2.30\qquad\qquad\therefore e^{2.3}\approx10\\ 
& \log 2 = 0.693\qquad\qquad\therefore e^{0.7}\approx2.
\end{align*}
And he could approximate small values by performing some mental math to get an accurate approximation to three decimal places. For example, approximating $e^{3.3}$, we have$$e^{3.3}=e^{2.3+1}\approx 10e\approx 27.18281\ldots$$But what I am confused is how Feynman knew how to correct for the small errors in his approximation. The actual value is$$e^{3.3}=27.1126\ldots$$Perhaps someone can explain it to me?
 A: As an enthusiast of mental math calculations,  here is what I think he did. Since he knew the basic approximations like you mentioned, he surely must have memorized the magnitude of the error: $$10-e^{2.3} = 0.02517...$$
But this is approximately $\dfrac{1}{4}\cdot 10^{-1}$. So basically it means you divide by $4$ and shift the decimal by one.  
Thus, he can see that a better approximation for $e^{3.3}$ would be $10e-\dfrac{e}{4}\dfrac{1}{10} = 27.1828... - 0.068.. = 27.1125.$ 
This method of mine takes literally couple of seconds and obtains an approximation almost the same as Feynman did, so perhaps he did something similar. 
There is no magic in mental calculations -  you just get used to seeing lots of numbers and figuring out how to deal with them. Like, it is better to divide by $4$ instead of multiplying by $0.25$, which is what I did above.  
A: According to the book (p. 124 here), he knew $\log 10 \approx 2.3026$. Feynman ostensibly used the trivial linear approximation of $e^x$:
$$e^x\approx 1+x$$
which works well for small values of $x$. 
Thus:
$$e^{3.3} = e^1\cdot e^{2.3026-0.0026} \approx e\cdot10\cdot e^{-0.0026}\approx 10e(1-0.0026) = 27.1121\dots$$
The correction adds two more correct decimal places and is quite easy to compute by hand.
Feynman used a similar trick to compute cubed roots faster than a man with an abacus (p. 127).
