Out of curiosity from this question regarding writing real numbers as an iterated sum/difference of square roots I started experimenting with another family of functions:

$$\log[ \exp[1] \pm (\exp[1/2]-1)\log[\cdots]]$$ or expressed as a recursive function with respect to the binary vector ${\bf b} = b_1b_2\cdots b_n, b_k \in \{0,1\}$: $$L({\bf b}) = \log[ \exp[1] + (-1)^{b_1-1} (\exp[1/2]-1)L[b_2b_3\cdots b_k]]$$

I am using the same idea as was proposed in the previous question ( solving one iteration "backwards" ) $$t_{n+1} = \frac{(\exp[t_n]-\exp[1]) }{ (\exp[1/2]-1)},\hspace{1cm} b_{n+1} = \text{sgn}(t_{n+1})$$ and then storing $\text{sgn}(t_{n+1})$ with the iteration of a new $\pm$ symbol on the inner log. I get this step-ladder behaviour when I plot the corresponding $x \rightarrow b$ below. Almost like some kind of rounding behaviour. While that is kind of exciting on it's own, my objective was to build a function with these $\pm$es able to represent all real numbers (on an interval) in the same sense as the radicals in the previous question could. As logarithms and exponentials are such smooth functions I figured it should be possible. Does anyone have a hint of how to go about?