       Re: Reduce[a^x + b^x - 2 == 0, x]

• To: mathgroup at smc.vnet.net
• Subject: [mg93655] Re: Reduce[a^x + b^x - 2 == 0, x]
• From: Bill Rowe <readnews at sbcglobal.net>
• Date: Thu, 20 Nov 2008 04:57:38 -0500 (EST)

```On 11/19/08 at 5:39 AM, severin at km1.fjfi.cvut.cz (Severin Posta)
wrote:

>Reduce[a^x + b^x - 2 == 0, x]

>into Mathematica 6.0.0 Win32, I get no result. System is running
>computation several days - probably forever :)

What is it you want to accomplish? What are you expecting?

Clearly, there are an infinite number of choices that will make
a^x + b^x - 2 == 0 True. For example, setting a=b=1 makes this
true for all x. Or if you want a slightly less trivial choice,
consider a=1/2, b= 3/2, x=1. More generally, b will be
(2-a^x)^(1/x) which can be obtained using Solve. So for any a <
1 you can find a x, b pair that will make the expression true.

Given all of this, what is it you want Reduce to do for you?

```

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