# 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…

#### Solutions Collecting From Web of "Proving that $2$ is the only real solution of $3^x+4^x=5^x$"

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.

If we insert the known solution we can write
$$5^{2+x} = 4^{2+x} + 3^{2+x}$$
asking, whether there might another solution exist besides $x=0$ . Then we can rewrite, putting the $5^x$ to the rhs:
$$5^2 = 4^2\cdot 0.8^x + 3^2\cdot 0.6^x$$
Then if the exponents $x$ on the rhs are zero, we have the known solution. But if $x$ increases over zero, then the values of both summands decrease simultaneously, thus the equality can no more hold.
The analogue occurs for decreasing $x$: both summands increase over their squares simultaneously, so there is no other solution possible. QED.