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?