Calculating logs and fractional exponents by hand In view of what we can compute by hand, on a piece of paper, without having to use a computer or a calculator, how far can we go with the evaluation of $\log$-functions and fractional powers?
More concretely, are there practical methods, that work in general well enough, for computing the followings?


*

*$\log(x)$ (natural log used here) for $x\in ]0,\infty[$ (e.g take $x=3,$ which has a natural log of $\approx 1.09$)

*$x^{\alpha},$ ($x$ being any real number) for $\alpha$ not being a natural number. So e.g. $\alpha=1/2,1/3,\cdots$


It would be neat to learn about possible ways of maybe first simplifying these calculations, translating them into more feasible calculations (by hand), and then take it from there. Or if direct methods exist that work for certain values or powers. This surely begs the question, how people used to do these calculations before computers were around.
 A: Before computers were available log tables were used to compute logs and fractional exponents. You say "by hand" but I'm assuming that reasonably sized pre-computed tables are allowed.
The method for estimating the log of an arbitrary number is as follows:
For example to compute $y=ln(14623)$ the first step is to find $log_{10}(14623)$.
$log_{10}(14623)= log_{10}(14.623 \times 10^3)= log_{10}(14.623)+3 $
The log tables are written for base 10 and a reasonable sized one can have results for $log_{10}(a)$ for all $a$ between 1 and 100.
We want to find $log_{10}(14.623)+3$. Round it to two significant figures of accuracy and find $log_{10}(15)+3$. 
Now we just look up $log_{10}(15)$ in the table which is $1.176$, therefore $log_{10}(14.623)+3 \approx 1.176+3=4.176$.
But we wanted the natural log so we use the change of base rule. $ln(x)=\frac{log_{10}(x)}{log_{10}(e)}$.
We know in advance that $log_{10}(e)=0.434$, therefore
$ln(14623)=\frac{4.176}{0.434}=9.616$
The method for computing exponents uses the above method.
To find $y=x^a$ first take the log of both sides
$log_{10}(y)=log_{10}(x^a)=alog_{10}(x)$
With the log tables we can find $log_{10}(x)$
With a table for exponents it is simple to estimate $y=10^{alog_{10}(x)}$
A: Let $a$ be the value you want to compute the natural logarithm of. Take the Taylor polynomial centered at $a$, e.g. $\ln(x)=\ln(a)+\frac{1}{a}(x-a)-\frac{1}{a^2}\frac{(x-a)^2}{2}+\frac{1}{a^3}\frac{(x-a)^3}{3}-\cdots$. We know that $\ln(1)=0$, so that 
$$\ln(a)\approx-\left(\frac{1}{a}(1-a)-\frac{1}{a^2}\frac{(1-a)^2}{2}+\cdots+\frac{1}{a^N}\frac{(1-a)^N}{N}\right)$$
for some $N$. 
Of course, if $a$ is far away from $1$, then $N$ may need to be very large. The series converges by the ratio test when $0<1<2a$, thus $\frac{1}{2}<a$ is required. Of course, if you require and $a$ below $0.5$, use the properties of logs: $\ln (2/7)=\ln2-\ln 7$.
By hand, this can become a monstrous calculation. For $a=3$, $N=17$ suffices to get accuracy up to $3$ decimal places, but for $a=25$, $N=100$ only gives $2$ decimal place accuracy. Not at all efficient, but it's a simple idea. It only involves basic algebraic operations and careful rounding.
To evaluate a general logarithm, just use the change of base. Of course, now you have to evaluate two natural logarithms, so this method becomes even more computationally intensive. 
To evaluate $b^a$, note that $b^a=e^{a\ln(b)}$, so we only need to know how to calculate exponents of $e$ and the natural logarithm (the latter is done above).
A similar Taylor series method as above gives
$$
e^a\approx \sum_{k=0}^N \frac{a^k}{k!}.
$$
The convergence properties of this approximation are pretty nice. Of course, you are still calculating integer powers of potentially unfriendly decimal expansions.
A: For back-of-the-envelope or mental computations, I just use $$\log_{10}(2)\approx0.3$$ (actually I remember $0.30103$). This implies $$\log_{10}(5)=0.7$$ I also recall that $\sqrt{10}\approx3.16$, so that $$\log_{10}(3.16)\approx0.5$$
From these values I grossly interpolate.
Examples:


*

*$\pi^{10}$: $\pi$ is close to $\sqrt{10}$, then even without logarithms, $10^5$ (vs. $93648.\cdots$).

*$\sqrt[5]{42}$: the log of $42$ will be a little more than $1+2\cdot0.3$, but let's stick to $1.6$. After division by $5$ we get $0.32$ corresponding to about the cube root of $10$, say $2.1$, as $2^3=8$ (vs. $2.112\cdots$).

*$e^{6.3}$: $e$ is $2.71828\cdots$ and its log is closer to $\sqrt{10}$ than to $2$, let us take $0.42$. Then the product is $2.646$, with antilog about $500$ (vs. $544.\cdots$).
What matters to me is a correct order of magnitude.
A: The well known series
$$\log(1+x) = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^k}{k}$$
converges for $-1< x\leq 1$, so it cannot be used to calculate $\log(3)$ with directly. However, it is possible to extend the radius of convergence using a conformal mapping. If we put $y = \frac{1+x}{1-x}$, then we have:
$$\log(y) = \log(1+x) - \log(1-x) =  2\sum_{k=0}^{\infty}\frac{x^{2k+1}}{2k+1}$$ 
Since $x = \frac{y-1}{y+1}$, this means that
$$\log(y) = 2\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(\frac{y-1}{y+1}\right)^{2k+1}$$ 
and this converges for all positive y. So, you can directly insert $y = 3$ in this series and compute $\log(3)$ quite accurately using only a few terms:
$$\log(3) =1 + \frac{1}{3\times 4} + \frac{1}{5\times 16} + \frac{1}{7\times 64} + \frac{1}{9\times 256} + \frac{1}{11\times 1024} +\cdots$$
So, with 6 terms we get 5 significant figures. But for larger $y$ the series will converge more slowly, it's then more convenient to use the above series to construct a series for $\log(y+1) -\log(y) = \log\left(1+\frac{1}{y}\right)$:
$$\log(1+y)  = \log(y)+ 2\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(\frac{1}{1+2y}\right)^{2k+1}$$ 
The series now converges faster when computing $\log(n)$ for $n$ larger than 2, but you then need to know $\log(n-1)$.  However, it is then possible to compute several logarithms simultaneously in terms of only fast converging series. E.g. to compute $\log(2)$, $\log(3)$ and $\log(5)$ simultaneously, we can use $2^4 = 16 = 15+1 = 3\times 5 +1$, $3^4 = 81 = 80+1 = 2^4\times 5 +1$, and $5^2 = 24+1 = 3\times 2^3+1$, this yields:
$$
\begin{split}
&4\log(2) - \log(3) - \log(5) &= 2\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(\frac{1}{31}\right)^{2k+1}\\
&4\log(3) - 4\log(2) - \log(5) &= 2\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(\frac{1}{161}\right)^{2k+1}\\
&2\log(5) - 3\log(2) - \log(3) & = 2\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(\frac{1}{49}\right)^{2k+1}
\end{split}
$$
So, you can then solve for the 3 logarithms using these 3 equations involving  fast converging series.
A: My favorite way of calculating the natural logarithm:
$$\ln(x)\approx-\gamma+\sum_{n=1}^x\frac1n$$
Where $\gamma$ is the Euler-Mascheroni constant.  Not only is this easy, but it works quite well for something like $x>10$.
If you need to calculate other bases, like $\log_b(x)$, use $\log_b(x)=\frac{\ln(x)}{\ln(b)}$.
If you are not satisfied with the result because $x$ is small, try $\ln(x)=\frac1k\ln(x^k)$ to make the approximation better.
A: Yes, it is possible.
For several hundred years, humanity has calculated log, ln, etc. by hand. It is a laborious, tedious task, and there is no easy way to verify the results either.
Even in the early 20th century, the results of these calculations were still printed and sold as books; any engineer had to have one.
