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?
 A: 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 $x=1$.
A: Let $f(x)=x^k,$ where $k>1$ or $k<0$.
Thus, $f$ is a convex function and since $(11,7,5)\succ(9,8,6),$ by Karamata we obtain:
$$f(11)+f(7)+f(5)>f(9)+f(8)+f(6).$$
Also, for $0<k<1$ we see that $f$ is a concave function.
Thus, by Karamata again
$$f(11)+f(7)+f(5)<f(9)+f(8)+f(6).$$
Thus, it remains to check, what happens for $k\in\{0,1\}$.
A: 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.
A: You write the equation in the form
$$
[(9+2)^k-9^k]-[(7+1)^k-7^k]-[(5+1)^k-5^k]=0.
$$
If $k\ge 0$, then you can use that 
$$
(9+2)^k-9^k=\int_9^{9+2}kt^{k-1}dt\ge 2k9^{k-1},\\
-[(7+1)^k-7^k]=-\int_7^{7+1}kt^{k-1}dt\ge -k8^{k-1},\\
-[(5+1)^k-5^k]\ge -k6^{k-1}.
$$
Your equation becomes
$$
0\ge k\cdot\left(2\cdot 9^{k-1}-8^{k-1}-6^{k-1}\right).
$$
If we discard the $k=0$ solution, it suffices to examine the inequality
\begin{align}
6^{k-1}+8^{k-1}&\ge 2\cdot 9^{k-1},\qquad k=1,2,3,4,...
\end{align}
This is equivalent to the inequality
$$
6^\ell+8^\ell\ge 2\cdot 9^\ell,\qquad \ell=0,1,2,3,...,
$$
if $k=\ell+1$.  Observe that
$$
9^\ell +9^\ell\ge 6^\ell+8^\ell\ge 2\cdot 9^\ell
$$
since $\ell\ge 0$.  This means equality holds, and the only way that can be true is if $\ell=0$.  (to see this step, rewrite as $0<9^\ell-6^\ell=-(9^\ell-8^\ell)<0$.)
