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3^x+4^x=5^x Conceptual Question

  1. Jul 14, 2009 #1
    I think I actually saw this question on here before but cant find it, hopefully someone can help me. We know x=2 for 3^x+4^x=5^x but how would you prove that x=2 in this case?


    Thanks.
     
  2. jcsd
  3. Jul 14, 2009 #2

    Borek

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    Staff: Mentor

    I am not sure what you are asking about.

    For x=2 9+16=25.

    Or do you mean how do we prove the equation has only one solution, x=2?
     
  4. Jul 14, 2009 #3
    I mean, how exactly would one go about proving x=2? If someone gave you the equation, or any other equation with an answer x=some number less then two how would you prove it?


    Basically, how would you solve for x, and show that x=2?
     
  5. Jul 14, 2009 #4
    Maybe solve graphically or use a numerical method?:P
     
  6. Jul 14, 2009 #5
    I'd rather not use a numerical method- my intent is to get an algebraic solution. As for graphically, I don't quite see how it would work. Does anyone know the correct analytical approach?
     
  7. Jul 14, 2009 #6
    I don't think there is an algebraic solution, but I could be wrong. To solve this graphically, plot [tex]f_1(x) = 3^x + 4^x[/tex] and [tex]f_2(x) = 5^x[/tex] on the same graph. Look for the point(s) where the two graphs intersect, where [tex] f_1(x) = f_2(x)[/tex]. Everywhere else, [tex]3^x + 4^x \neq 5^x[/tex].
     
  8. Jul 16, 2009 #7
    I don't see how this is a problem. We can go back to 2 = 1+1, 2^2 = (1+1)+(1+1) = 2(1+1), and go on with this and count the 'sticks', arriving at a 1 to 1 correspondance. This is how Cantor decided two sets were the same.

    If you are worried about whether there are other solutions for higher x, we have the inequality 3^x+4^x < 5^x for x>2.
     
    Last edited: Jul 16, 2009
  9. Jul 16, 2009 #8
    I'm fairly certain there is no algebraic method. If there was some algebraic way to get x as a function of {3,4,5} then we could replace {3,4,5} with (say) {a,b,c} and have an analytical solution for the general equation. But for a,b,c,x integral, we have Fermat's Last Theorem which clearly does NOT have a simple solution (or indeed any solution).

    Edit: To clarify, what I am saying is that if we could find some analytical solution to the above problem then from this solution it would surely then be possible to prove that for a,b,c,x integral and x>2 there is no solution i.e. prove Fermat's Last Theorem. Since the proof of this theorem is decidedly nontrivial (!), it seems very unlikely that there is an analytical solution!
     
    Last edited: Jul 16, 2009
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