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

[SHawking]

I think I actually saw this question on here before but can't 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.

 

[Borek]

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?

 

[SHawking]

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?

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?

 

[BobbyBear]

Maybe solve graphically or use a numerical method?:P

 

[SHawking]

Maybe solve graphically or use a numerical method?:P

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?

 

[Fenn]

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?

I don't think there is an algebraic solution, but I could be wrong. To solve this graphically, plot f_1(x) = 3^x + 4^x and f_2(x) = 5^x on the same graph. Look for the point(s) where the two graphs intersect, where f_1(x) = f_2(x). Everywhere else, 3^x + 4^x \neq 5^x.

 

[robert Ihnot]

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.

 

[Alewhey]

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!