Proving that $ 2 $ is the only real solution of $ 3^x+4^x=5^x $ 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'(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...

One direct method is to divide directly by $5^x$ and get $1=(3/5)^x+(4/5)^x$. From here it is clear that the RHS is strictly decreasing, and there is a unique solution. Almost all exponential equations can be treated this way, by transforming them to


*

*one increasing function equal to one decreasing function

*one increasing/decreasing function equal to a constant.
