       Re: Overly complicated reductions?

• To: mathgroup at smc.vnet.net
• Subject: [mg77674] Re: Overly complicated reductions?
• From: dh <dh at metrohm.ch>
• Date: Thu, 14 Jun 2007 06:20:01 -0400 (EDT)
• References: <f4om08\$7bj\$1@smc.vnet.net>

```
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 \[Element] Integers && y != 0 &&

> x == 1\/2\ \((2\ \[ImaginaryI]\ \[Pi]\ C + 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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