       Re: Overly complicated reductions?

• To: mathgroup at smc.vnet.net
• Subject: [mg77638] Re: [mg77584] Overly complicated reductions?
• From: DrMajorBob <drmajorbob at bigfoot.com>
• Date: Thu, 14 Jun 2007 05:26:22 -0400 (EDT)
• References: <22445676.1181737009723.JavaMail.root@m35>
• Reply-to: drmajorbob at bigfoot.com

```Reduce gave you ALL the solutions. This narrows it down to the one you
want:

Reduce[y == E^(-2 x), x]
% /. C -> 0
ToRules[%]
x /. %

C \[Element] Integers && y != 0 &&
x == 1/2 (2 \[ImaginaryI] \[Pi] C + Log[1/y])

y != 0 && x == 1/2 Log[1/y]

{x -> 1/2 Log[1/y]}

1/2 Log[1/y]

Now we can test to see if that matches your posted answer.

Log[y]/-2 == 1/2 Log[1/y] // Simplify
% // PowerExpand

Log[1/y] + Log[y] == 0

True

I used PowerExpand to get Mathematica to assume y is a positive real
number. That avoids the very same branch-cut issues that gave your
original problem multiple answers.

Bobby

On Wed, 13 Jun 2007 06:36:03 -0500, David Rees
<w3bdevilREMOVE at THISw3bdevil.com> 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
>
>
>
>

--
DrMajorBob at bigfoot.com

```

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