Iterative calculation of $\log x$ Suppose one is given an initial approximation of $\log x$, $y_0$, so that:
$$y_0 = \log x + \epsilon \approx \log x$$

Here, all that is known about $x$ is that $x&gt;1$. Is there a general method of improving that estimation using only addition &amp; multiplication, i.e. without exponentiation or logarithms?
$$y_1 = f(y_0, x)=\ ?$$
Instead of solving $y - \ln(x) = 0$ for $y$ you can solve $g(y) = e^y - x = 0$

Given initial approximation $y_0 \approx \ln(x)$ you can try to solve $y$ using Newton's medhod:

$$y_{n + 1} = y_n - \frac{g(y_n)}{g'(y_n)}= y_n - \frac{e^{y_n} - x}{e^{y_n}} = y_n - 1 +\frac{x}{e^{y_n}}$$

When exponentiation isn't allowed you can approximate $e^{y}$ with $\left(1 + \frac{y}{2^m} \right)^{2^m}$ using repeated multiplication: $$\begin{aligned} \left(1 + \frac{y}{2^m} \right)^{2^m} &amp;= \left(1 + \frac{y}{2^m} \right)^{2^{m - 1}} \cdot \left(1 + \frac{y}{2^m} \right)^{2^{m - 1}} \\ \\2^{-m} &amp;= \frac{1}{2} \cdot 2^{-m + 1}\end{aligned}$$
A long time ago, in the last century, I had an early calculator with no log function.  To do a log, I would hit the square root button ten times, subtract $1$, and multiply by $1,024$.  The answer was good to $4$ decimal places, which was better than my two calibrated sticks could deliver.  Square root is a powerful function that can be used to approximate many others.

Do not ridicule calibrated sticks; they did most of the engineering that got us to the moon!  I recently built a slide rule from Starbucks sticks that does multiplication modulo $23$.  I used logs to base $7$.