Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$ 
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$

It's easy to see that $x=0$ and $x=1$ are solutions but are these the only one? How do I demonstrate that?
I've tried to write them either:
$$5^x+7^x+11^x=2^x*3^x+2^{3x}+3^{2x}$$
or
$$5^x+7^x+11^x=(5+1)^x+(7+1)^x+(11-2)^x$$
and tried to think of some AM-GM mean inequality or to divide everything by $11^x$, but those don't seem like the way to go. Any hints?

Consider the function for $x>1,n>1$
$$f(n)=n^x-(n-1)^x$$
$$f'(n)=xn^{x-1}-x(n-1)^{x-1}=x(n-1)^{x-1}[(1+\frac{1}{n-1})^{x-1}-1]>0$$
So $f(n)$ is increasing for $x>1,n>1$.
Now rewrite the equation
$$5^x+7^x+11^x=6^x+8^x+9^x$$
$$\color{red}{(11^x-10^x)}+\color{blue}{(10^x-9^x)}=\color{red}{(8^x-7^x)}+\color{blue}{(6^x-5^x)}$$
Comparing the red and blue parts, LHS is larger than RHS due to increasing $f(n)$. 
So the equation holds only if $0\le x\le 1$ for natural number solutions, i.e. $x=0$ or $x=1$.
