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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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