Perhaps this is a simple question but I can't see a way to understand it. I have some expression of the form ##\ln(1-1/(1-x))## and I perform the following manipulations using simple algebra and laws of logs:(adsbygoogle = window.adsbygoogle || []).push({});

$$\ln \left(1-\frac{1}{1-x}\right) = \ln \left(\frac{1-x-1}{1-x}\right) = \ln \left(\frac{x}{(x-1)}\right) = -\ln \left(\frac{x-1}{x}\right) = -\ln \left(1-\frac{1}{x}\right)$$

I am doing such manipulations under the premise of ##x \in (0,1)##. However, if I put x equal to 1/2 then the left hand side does not match the right hand side. It does for x otherwise though. So, is it correct that the above rewriting is only valid for x not 1/2 and ,if so, why and how to rewrite ##\ln(1-1/(1-x))## in terms of ##\ln(1-1/x)## valid for ##x \in (0,1)##?

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# I Validity of laws of logs

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