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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[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
>
>
>

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






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