Iterative calculation of log⁡x Suppose one is given an initial approximation of log⁡x,y0, so that:...

Instead of solving ${\displaystyle y-\ln \left(x\right)=0}$ for y you can solve ${\displaystyle g\left(y\right)=e^y-x=0}$
Given initial approximation ${\displaystyle y_0\approx \ln \left(x\right)}$ you can try to solve y using Newtons

Consider the fact that the number you are working with is a floating point number, then the log of it is
${\displaystyle \log \left(b\right)=\log (2^{\exp }\cdot 1.S)}$, where exp is the exponent and S is the significand.
${\displaystyle \log \left(b\right)=\exp \cdot \log \left(2\right)+\log (1.S)}$
Since 1.S is between 1 and 2, assume it to be 2 then
${\displaystyle \log \left(b\right)\sim \left(\exp +1\right)\cdot \log \left(2\right).}$
a couple iterations should give the correct value then.