Re: Overly complicated reductions?

*To*: mathgroup at smc.vnet.net*Subject*: [mg77638] Re: [mg77584] Overly complicated reductions?*From*: DrMajorBob <drmajorbob at bigfoot.com>*Date*: Thu, 14 Jun 2007 05:26:22 -0400 (EDT)*References*: <22445676.1181737009723.JavaMail.root@m35>*Reply-to*: drmajorbob at bigfoot.com

Reduce gave you ALL the solutions. This narrows it down to the one you want: Reduce[y == E^(-2 x), x] % /. C[1] -> 0 ToRules[%] x /. % C[1] \[Element] Integers && y != 0 && x == 1/2 (2 \[ImaginaryI] \[Pi] C[1] + Log[1/y]) y != 0 && x == 1/2 Log[1/y] {x -> 1/2 Log[1/y]} 1/2 Log[1/y] Now we can test to see if that matches your posted answer. Log[y]/-2 == 1/2 Log[1/y] // Simplify % // PowerExpand Log[1/y] + Log[y] == 0 True I used PowerExpand to get Mathematica to assume y is a positive real number. That avoids the very same branch-cut issues that gave your original problem multiple answers. Bobby On Wed, 13 Jun 2007 06:36:03 -0500, David Rees <w3bdevilREMOVE at THISw3bdevil.com> 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 > > > > -- DrMajorBob at bigfoot.com