# Is this an equivalent uniform distribution?

Suppose $$U$$ is uniformly distributed on $$(0,1)$$ then it is true that $$X=\ln(1-U)\equiv \ln(U)$$. This makes sense because $$1-U\sim U$$, and thus $$X$$ will have the same distribution. Now, suppose $$c\in(0,1)$$, then why can't I say that $$\ln(1-(1-c)U)\equiv\ln((1+c)U)$$ since $$1-(1-c)U=1-U+cU$$, and given that $$1-U\sim U$$, $$1-U+cU=U+cU=(1+c)U$$. Thus $$1-(1-c)U\sim(1+c)U?$$

If $$X \sim Y$$ and $$X' \sim Y'$$ we cannot say $$X+X' \sim Y+Y'$$ without some independence assumptions. In this case $$1-U$$ and $$cU$$ are not independent (unless $$c=0$$) so the argumnet fails.