Calculating logs and fractional exponents by hand

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.

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.

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 $y=10^{alog_{10}(x)}$