TITLE: How did feynman compute $e^x$ with the accuracy he wanted?
QUESTION [5 upvotes]: Excerpt from the book Surely You're Joking, Mister. Feynman!

Here Feynman calculates $e$ to a couple of powers.
I understand that he luckily knew a couple of logs by heart. What I don't understand is the part where he adjusts the numbers to the accuracy he wanted to have. The accuracy he got in the first calculation e^3.3 was 27.1126.
In the first example it could be : $$(e^{2.3026} \times e) / e ^ {0.0026} = e^{3.3}$$ 
But then he still had to use the power series for the $e^{0.3026}$ AND divide by it afterwards and I can't see how even Feynman would be able to do that.
Maybe by a linear approximation? Thanks in advance.

REPLY [2 votes]: $e^{3.3} = e^{2.3026\dots} \times e \times e^{-0.0026\dots} = 10 \times e \times e^{-0.0026\dots}$. At this point you could use the linear approximation I guess,
$$
  e^{-0.0026\dots} =1 - 0.0026{\dotsc} + \frac{0.0026{\dotsc}^2}{2} - \dotsb \approx 1-0.0026{\dotsc} \approx 0.9974 .
$$
I don't know if this is what Feynman actually did.