This calculator takes the base e logarithm of a number.
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In general, there isn't an exact way to do this. Manually approximating the log of a number is hard! To make it easier, we must remember three things. Firstly, that \begin{equation} \label{series} \tag{1} \ln(x) = 2 \left( k + \frac{k^3}{3} + \frac{k^5}{5} + \ldots \right) \qquad \text{where } k = \frac{x-1}{x+1}, \end{equation} secondly, \begin{equation} \ln(10) \approx 2.30 \qquad \text{(3 significant figures)}, \end{equation} and lastly we must remember our log laws!
Now, suppose we want to find an approximation of \(\ln(x_0)\), then we would rewrite it as \begin{equation*} \ln(x_0) = \ln(x_0/10^n) + n \times \ln(10) \qquad \text{where } \frac{x_0}{10^n} \approx 1. \end{equation*} After rewriting it we can figure out \(\ln(x_0/10^n)\) using equation \(\eqref{series}\), and we know that \(\ln(10) \approx 2.30\). The reason we want \(x_0/10^n\) to be close to \(1\) is so that we can use fewer terms from equation \(\eqref{series}\) to get a good approximation (faster convergence).
As an example, if we want to find \(\ln(643)\) we would do the following \begin{align*} \ln(643) &= \ln(0.643 \times 10^3) \\ &= \ln(0.643) + \ln(10^3) \\ &= \ln(0.643) + 3 \ln(10) \\ &\approx \ln(0.643) + 3 \times 2.3. \tag{2} \label{approx1} \end{align*}
To get \(\ln(0.643)\) using \(\eqref{series}\), we write \(k = (0.643 - 1)/(0.643 + 1) \approx -0.217 \) so that \begin{align*} \ln(0.643) & \approx 2 \left( -0.217 + \frac{(-0.217)^3}{3} + \frac{(-0.217)^5}{5} \right) \\ & \approx -0.441. \tag{3} \label{approx2} \end{align*} Notice that we've only had to write out the first three terms of equation \(\eqref{series}\) to get an acceptable approximation in this case.
Substituting the result from equation \(\eqref{approx2}\) back into \(\eqref{approx1}\) we get \begin{align*} \ln(643) & \approx -0.441 + 3 \times 2.30 \\ & \approx 6.5, \end{align*} which is pretty close! In fact, using the calculator on this page we get \(\ln(643) \approx 6.46614\).