I came across the following question:  Find the number of solutions of the equation $3^x+4^x=5^x$ in the set of positive real numbers.  I tried the above question by taking log on both sides and then solving, but it didn't seem to work. Any idea on how to proceed to solve this question?
I would like to prove that the equation $ 3^x+4^x=5^x $ has only one real solution ($x=2$)  I tried to study the function $ f(x)=5^x-4^x-3^x $ (in order to use the intermediate value theorem) but I am not able to find the sign of $ f&#39;(x)= \ln(5)\times5^x-\ln(4)\times4^x-\ln(3)\times3^x $ and I can't see any other method to solve this exercise...
We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($x\neq y$) which satisfies the equality $x^y = y^x$?