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Re: Overly complicated reductions?
On 13 Jun 2007, at 20:36, 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 > > > But Mathematica is completely right (over the complex numbers) and your by hand solution is just a very special case. To get it use: Reduce[Simplify[Reduce[y == E^(-2*x), x, Reals], y > 0], x] x == -(Log[y]/2) Andrzej Kozlowski