I can do this using the following relation:
$$\log(1 - x) = \log(1 - \exp(y))$$
Here $y = \log(x)$ is always a negative number. However, I was wondering whether it's possible to compute $\log(1 - x)$ without using exponentiation.
2026-04-03 00:20:52.1775175652
How to evaluate $\log(1 - x)$ in terms of $\log(x)$?
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COMMENT.-It is true @geodude's comment. Anyway let $f(x)=\log(1-x)$ and $g(x)=\log(x)$. If you make the change of variable $\frac 12+x$ instead of $x$ then you have $f(X)=\log(\frac 12-X)$ and $g(X)=\log(\frac 12+X)$ in which the symmetry of the functions with respect to the line $x= \frac12$ becomes evident and $f(X)=g(-X)\iff f(-X)=g(X)$. Can you use this?