I have a function
$$\ f(x)=3^x+4^x-5^x $$
How to prove taht f(x) has only one solution in real numbers? I have tried to take a derivative through $\ e^{x\ln3} $ substitution:
$$\ f(x)' = \ln(3)3^x+ \ln(4)4^x- \ln(5)5^x $$
However, I am stuck at the moment and I don't know how to continue.
On
This is equivalent to finding the solutions of
$$\left(\frac35\right)^x+\left(\frac45\right)^x=1.$$
But the derivative of the LHS is
$$\log\frac35\left(\frac35\right)^x+\log\frac45\left(\frac45\right)^x<0$$ and the function is strictly decreasing.
As $x=2$ is a solution, it is the only one.
For a very rough solution, notice that $0.6\approx 0.8^2$ and with $t:=0.8^x$,
$$t^2+t=1\implies t=\phi-1=0.618\cdots$$ and $$x\approx\log_{0.8}(\phi-1){}\approx 2.15\cdots$$
This can be used as a starting value for Newton's iterations.
Hint: Consider $\frac{f(x)}{4^x}$.