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Re: Overly complicated reductions?
David Rees wrote: > Consider f(x)=e^(-2x) > > I wanted to retreive the inverse function f^-1(x), Mathematica to the > rescue: > \!\(Reduce[y == E\^\(\(-2\) x\), x]\) > > \!\(C \[Element] Integers && y != 0 && > x == 1\/2\ \((2\ \[ImaginaryI]\ \[Pi]\ C + Log[1\/y])\)\) > > This can't be right, I can rearrange it to just Ln(x)/-2 on paper. What did > I do wrong? > > Thanks You could specify a domain to get something closer to what you are looking for. For instance, In:= Reduce[y == E^(-2*x), x, Reals] Out= y > 0 && x == (1/2)*Log[1/y] Regards, Jean-Marc