Re: Overly complicated reductions?

*To*: mathgroup at smc.vnet.net*Subject*: [mg77674] Re: Overly complicated reductions?*From*: dh <dh at metrohm.ch>*Date*: Thu, 14 Jun 2007 06:20:01 -0400 (EDT)*References*: <f4om08$7bj$1@smc.vnet.net>

Hi David, Reduce gives the correct answer. If you do not belive it, simply replace x in Exp[-2x] by the solution and you will get y. Obviously for y=0 there is no solution (or infinity if you like) 2Pi n comes in because Exp[2PI n]==1 If you only want a real solution, replace y==.. by {y==..,Element[x,Reals]} hope this helps, Daniel 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[1] \[Element] Integers && y != 0 && > x == 1\/2\ \((2\ \[ImaginaryI]\ \[Pi]\ C[1] + 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 > > >