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?

Another elemenatary solution, using that $x$ is supposed to be a natural number. It even uses a technique the OP considered:
The equation is equivalent to
$$\left(\frac{5}{11}\right)^x + \left(\frac{7}{11}\right)^x +\left(\frac{11}{11}\right)^x = \left(\frac{6}{11}\right)^x +\left(\frac{8}{11}\right)^x +\left(\frac{9}{11}\right)^x$$
All terms are positive and the left hand side contains a summand 1 in the form of $\left(\frac{11}{11}\right)^x$.
From the terms on the right hand side, $\left(\frac{9}{11}\right)^x$ is the biggest, but it will of course still decrease for increasing $x$. Using a calculator will show you that $\left(\frac{9}{11}\right)^6 < \frac13$.
That means for $x \ge 6$, the right hand side consists of the sum of 3 values, the highest of which is less than $\frac13$. That means the right hand side is less than 1, while the left hand side is bigger than 1, leading to a contradiction.
Now 'only' the cases $x=2,3,4,5$ need to be checked by hand, and they don't lead to the equation being fullfilled.
