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

*To*: mathgroup at smc.vnet.net*Subject*: [mg93678] Re: Reduce[a^x + b^x - 2 == 0, x]*From*: Scott Hemphill <hemphill at hemphills.net>*Date*: Fri, 21 Nov 2008 05:34:44 -0500 (EST)*References*: <gg3cab$k5u$1@smc.vnet.net>*Reply-to*: hemphill at alumni.caltech.edu

Bill Rowe <readnews at sbcglobal.net> writes: > 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? Let me guess: given a constant a > 0 and b > 0, he would like a formula for the non-trivial real value of x which satisifies this equation. The closest I have come is to note that if a and b are rational powers of each other, x can be expressed in terms of the root of a polynomial. Mathematica can calculate this if you give it specific examples to work with: In[1]:= a = 2; b = a ^ (-2/3); In[2]:= Reduce [a^x+b^x==2, x, Reals] 2 3 4 3 Log[Root[-1 - #1 + #1 + #1 + #1 & , 2]] Out[2]= x == 0 || x == -------------------------------------------- Log[2] Scott -- Scott Hemphill hemphill at alumni.caltech.edu "This isn't flying. This is falling, with style." -- Buzz Lightyear