> Log(-1) is immediately replaced by %i*%pi.
> There is then only one log in the expression.
===========
> (%i1) assume(a<1);
> (%o1) [a<1]
> (%i2) log(a-1) - log(-1);
> (%o2) log(a-1)-log(-1)
> (%i3) rectform(%);
> (%o3) log(1-a)
Great! That's the trick. But one has to know Richard Fateman's
explanation in order to understand why this works. Thanks to both of you!
Wolfgang Hugemann