how to prove 3^x + 4^x = 5^x?

Question

Its,
3 power x + 4 power x = 5 power x
And please don't use that hit and trial method to give value of x = 2.

Answer 1

You problably mean: how to *solve* 3^x + 4^x = 5^x?

There is no analytical way of doing it, but you can find the solution by iteration. Rewrite the equation as x = ln (3^x + 4^x)/ln 5.
Take a starting value for x, put in the rhs, and obtain a new x=2.80...
Plug in again. Continue iteration like that and the x will slowly converge to x=2.

Answer 2

I agree with the other posters: I believe you mean you want to find all x such that 3^x + 4^x = 5^x.

If you divide both sides by 5^x then you want to solve the equivalent equation (3/5)^x + (4/5)^x = 1. I'm going to prove that the function f(x) = (3/5)^x + (4/5)^x - 1 vanishes at only one point, namely x = 2. This is obvious if you plot f(x) vs. x, but I'll prove this rigorously.

I claim that f(x) is a strictly decreasing function. (Again, this is obvious if you stare at the plot of f(x).) This is because it's derivative is f'(x) = ln(3/5)*(3/5)^x + ln(4/5)*(4/5)^x, the logarithm ln(3/5) < 0, yet the exponential (3/5)^x > 0. Hence f'(x) < 0 for all x.

This strictly decreasing function only has at most one value x such that f(x) = 0. (More generally, there is at most one x such that f(x) = A for any given A.) Now f(2) = 0 so x = 2 is the only such value such that f(x) = 0.

Answer 3

Omg, all the answers are so complicated!!!
Actually I'm allittle bit dumb at the moment, so I can't give you 'pure' algebraical solution, but I have another one.
OK then:
3^x + 4^x = 5^x
(3/5)^x + (4/5)^x = 1
(3/5)^x = 1 - (4/5)^x
We have to functions now, which can have only one point of intersection. That can be proven by doing some mathematical analysis craft, I won't post it here :P
You don't like a "hit and trial method", but 3, 4, 5 are well-known Pythagorean numbers that satisfy a right-angled triangle equality
a^2+b^2=c^2. So you don't just try a random number here. You know that x=2 is the solution, your mission is to prove it's uniqueness--that has been done above.
I hope you accept my answer.

Answer 4

x=0

Answer 5

If x=4, 3^4 + 4^4 does not equal 5^4, so your equation does not hold for all x.

Answer 6

That is actually very similar to Fermat's Last Theorem, in which he states there are no values greater than 2 for which the equation A^x+B^x= C^x holds true.

So in order to prove what you are asking, you need to ask Andrew Wiles

Answer 7

It is only true for x=2. Pythagorean theorem.

Answer 8

use natural log (ln) function

Answer 9

http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem#The_proof_of_Fermat.27s_Last_Theorem


Doug

Answer 10

it is dream