$\displaystyle 3^x+4^x=5^x$ Show that the equation $\displaystyle 3^x+4^x=5^x$ has exactly one root
I proved it graphically but I am in need of an analytic solution
 A: HINT:
Observe that $\frac35,\frac45<1,$
So, $$\left(\frac35\right)^m+\left(\frac45\right)^m<\left(\frac35\right)^2+\left(\frac45\right)^2=1  \text{  for }m>2$$
and $$\left(\frac35\right)^m+\left(\frac45\right)^m>\left(\frac35\right)^2+\left(\frac45\right)^2=1 \text{  for }m<2$$
This can be generalized for any  Pythagorean triple
A: $3^x+4^x=5^x\iff\left(\frac35\right)^x+\left(\frac45\right)^x=1$. Since the exponential function on the left hand side is strictly decreasing, it follows that the equation, were it to ever be true for a certain x, then that x is unique. And it is indeed true for $x=2$, making it its only solution.
A: A hint: Use that
$$x\mapsto\left({3\over5}\right)^x+\left({4\over5}\right)^x=e^{-\log\!{5\over3}\>x}+e^{-\log\!{5\over4}\>x}$$
is strictly decreasing on ${\mathbb R}$.
A: Hint: Use the Intermediate Value Theorem to show that $f(x) = 3^x+4^x-5^x$ has a zero on some interval.  The use Rolle's theorem to show you can't have more than one zero.
A: 
write 5^x=(4+1)^x and use binomial expansion 
then simplify 
it seems to have answer  only for x=2 
