MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

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[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

You could specify a domain to get something closer to what you are 
looking for. For instance,

In[1]:= Reduce[y == E^(-2*x), x, Reals]

Out[1]= y > 0 && x == (1/2)*Log[1/y]

Regards,
Jean-Marc


  • Prev by Date: Background Option in LocatorPane
  • Next by Date: Re: Axis missing from simple ListPlot
  • Previous by thread: Re: Overly complicated reductions?
  • Next by thread: Re: Overly complicated reductions?