Pretend you are in 1859. What is a fast, efficient, and accurate way to numerically evaluate constants like that to, say, 20 decimal places, using ONLY pen and paper?
Answer
The French paper of Hermite (1859) ‘Sur la théorie des equations modulaires’ is available freely at Google books, it begins at page 29.
You’ll find there the value of eπ√43 with all the correct digits but concerning eπ√163 there is only the indication that the fractional part should begin with twelve consecutive 9. He clearly used modular properties to deduce this as explained by Matt. Wikipedia’s page concerning Heegner numbers could help too.
At first glance Hermite doesn’t seem to provide his computations’ secrets concerning eπ√43, polynomials of order 48 of page 67 (he speaks of “quite long but not at all impractical calculation” about them) and so on… Clearly he didn’t fear tedious computations!
So let’s see what we can do with logarithms or exponential tables. The ‘state of the art’ for these (1859) days may perhaps be found in this paper of 1881 from Alexander Ellis : Gray published Tables for twelveplace logarithms in 1845 and Thoman a “Tables de logarithmes à 27 décimales” in 1867. Twelve digits seem enough for the nine digits of the integer part of eπ√43 and Hermite probably deduced the fractional part by inversion and multiplication of this result.
Let’s suppose arbitrary that he used exponential tables (he could as well have used logarithm tables or log10 tables or something more subtle ‘AGM like’ who knows…)
I’ll start with π√43≈20.6008006943 (√43 is obtained quickly by iterations of n′=n2+43d2,d′=2nd) and try to get eπ√43≈884736744.000
I used the ‘relatively recent’ Abramowitz and Stegun tables that I’ll ’round’ at 12 digits for my ‘Hemulation’.
e20 is tabulated page 138 as ( 8)4.85165 19540 98 and this value of 485165195.410 will be our reference. We will have to multiply this by e0.6008006943 (nearly 1.82357834477). Let’s try to get this one :

by interpolation between e0.600≈1.82211880039 and e0.601≈1.82394183055 from the tables we get :
1.82357849025 with the final result 884736814.567 clearly insufficient…
To get something more exact we need more tabulated values or expand e0.601−0.6008006943 in Taylor series getting e0.601e−0.0002e0.0000007e−0.000000005=1.82357834605 with the final result 884736744.607 not to far from our target (the idea is that the Taylor series for e−0.0002, e0.0000007… are fast) 
or by multiplication by e0.6, e0.0001 (8 times), e0.0000001 (6 times), e0.00000001 (9 times), e0.000000001 (4 times) at this point the final result is 884736743.722 and we are nearly there…
Let’s note that the second method allows to get high precision just by ‘preevaluation’ of some terms. In fact e10k (for k=1,0,−1,−2⋯−18) evaluated with high precision should have worked for eπ√163 !
EDIT2: evaluating directly e0.0000006, e0.00000009, e0.0000000043 to 18 digits would probably be more efficient and the result could be evaluated this way : aeϵ=a+(a)ϵ1+(aϵ1)ϵ2+⋯ (each time the previous term is mutiplied/divided by ϵn and added to the result)
some excellent mental calculators of these days could have contributed too!
Of course Hermite’s method could have be much more subtle. Vladimir Arnold pointed out that many techniques like elliptic functions were better known in the nineteenth century than now. I think that the same could be said about quite some methods to solve Diophantine equations before the time of computers as you may confirm! 🙂
OTHER METHODS : Let’s try to apply the general method proposed by Apostol (in tzs’ answer) or rather the idea of another Caltech professor for computing ‘exponentials in your head’ : Richard Feynman (see J.M.’s extract ‘Lucky Numbers’ from Feynman’s very funny book “Surely You’re Joking, Mr. Feynman!”).
Let’s look at one of his favorite examples : he was asked to compute e3 ‘in his head’ and was able to answer quickly 20.085. He needed only the values of ln2 and ln10 in this case. Let’s detail this (to 6 digits here) : ln10≈2.302585 and ln2≈0.693147 but their sum is ≈2.995732=3−ϵ with ϵ≈0.004268 so that :
e3=eln10+ln2+ϵ≈10⋅2(1+ϵ(1+ϵ2))≈20+0.08536+0.000182≈20.085542
If asked for e4 he would simply multiply this by 2.7182818 to get : e4≈54.59816 (the relation used is 4≈1+ln2+ln10).
A generalization of Feynman’s trick to compute ea would consist in searching linear relations between a, 1, ln2 and ln10 (we may accept other constants especially logarithms since ec should be easy to evaluate). This is easier to do now with algorithms like PSLQ or LLL, in Hermite’s days I think you could only use continued fractions or guess…
For π√43 we may get excellent approximations like 218153, 564/5.
For π√163 approximations are 230⋅512, 210⋅37⋅511⋅74, 27⋅320⋅5⋅76, e404⋅38⋅5/76 (I suppose that we have a table of logarithms and of exponentials of integers given with high precision).
But to be honest I think that none of these tricks were used by Hermite himself for these evaluations because he knew, as you may find in the previous links, ‘Heegner numbers’ for example, that the nearest integer was given by 9603+744=884736744 for eπ√43 and by 6403203+744=262537412640768744 for eπ√163 so that he needed no evaluation of exponential at all !!
Attribution
Source : Link , Question Author : Tito Piezas III , Answer Author : Community