Re: Overly complicated reductions?
- To: mathgroup at smc.vnet.net
- Subject: [mg77657] Re: Overly complicated reductions?
- From: dimitris <dimmechan at yahoo.com>
- Date: Thu, 14 Jun 2007 05:36:07 -0400 (EDT)
- References: <f4om08$7bj$1@smc.vnet.net>
In[1]:= Reduce[y == E^(-2*x), x] Out[1]= C[1] Integers && y != 0 && x == (1/2)*(2*I*Pi*C[1] + Log[1/y]) In[3]:= PowerExpand[Reduce[y == E^(-2*x), x, Reals]] Out[3]= y > 0 && x == -(Log[y]/2) In[7]:= PowerExpand[Reduce[y == E^(-2*x) && 0 < x < 1, x]] Out[7]= 1/E^2 < y < 1 && x == -(Log[y]/2) Dimitris / David Rees : > 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