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Re: Overly complicated reductions?



Hi David,

Reduce gives the correct answer. If you do not belive it, simply replace 

x in Exp[-2x] by the solution and you will get y.

Obviously for y=0 there is no solution (or infinity if you like)

2Pi n comes in because Exp[2PI n]==1

If you only want a real solution, replace y==.. by {y==..,Element[x,Reals]}

hope this helps, Daniel



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

> 

> 

> 




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