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Re: Overly complicated reductions?


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



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