How to show that $3^x+4^x=5^x$ has only one solution? How to show that $3^x+4^x=5^x$ has only one solution? Thanks in advice.

Define
$$f(x)=\left(\dfrac{3}{5}\right)^x+\left(\dfrac{4}{5}\right)^x.$$ 
It is clearly monotonic decreasing. Note also that
$$f(2)=1,$$
so that
$$x=2$$ is the only real solution to $f(x)=1$.
More generally, for the equation
$$3^x+4^x+5^x=6^x,$$
define
$$f(x)=\left(\dfrac{3}{6}\right)^x+\left(\dfrac{4}{6}\right)^x+\left(\dfrac{5}{6}\right)^x.$$
This function is also decreasing, and as Euler first noted,
$$3^3+4^3+5^3=6^3,$$
so that
$$f(3)=1,$$
and $x=3$ is the only real solution to $f(x)=1$.
