Find all real number satisfying $10^x+11^x+12^x = 13^x+14^x$ 
Find all real number $x$ satisfying $10^x+11^x+12^x = 13^x+14^x$    

My Work
Dividing by $13^x$ we get 
$$\left( \frac{10}{13} \right)^x + 
\left( \frac{11}{13} \right)^x + 
\left( \frac{12}{13} \right)^x
= 1 + \left( \frac{14}{13} \right)^x$$
The LHS is a decreasing function of $x$ and the RHS is an increasing function of $x$. So there is only one intersection in their graph. I am looking for a formal way to find the root. I know that $x=2$ works. But how to formally find this root?

If you know that the solution must be an integer, this type of equation is known as exponential Diophantine, and there is no known formal procedure to solve it in the general case and probably none exists (this was proven for ordinary Diophantine equations).
If the solution is allowed to be real, there is no systematic procedure either as this is a transcendental equation. You need to resort to numerical methods to estimate the roots, and this requires a step of root separation (finding intervals that are guaranteed to contain exactly one root). Unfortunately, root separation can require the resolution of an even more difficult equation to get the extrema. In the given case, you are lucky as the function is easily shown to be monotonic.
After root isolation, you can evaluate the root to arbitrary precision (at least in theory) using some unidimensional root solver (dichotomy, secant, Newton...)
If it turns out that the root seems to have a simple form (integer, rational or some other closed expression), it may possible to prove or disprove it formally, but here again, no systematic method.
In the given case, this is straightforward:
$$11^2+12^2+13^2=365=14^2+15^2.$$
