Solve $10^x+11^x+12^x = 13^x+14^x$ 
Find all real numbers $x$ for which $$10^x+11^x+12^x = 13^x+14^x.$$

I thought about taking the logarithm of both sides, but didn't see how that would help. How can we make the equation simpler?
 A: As @CatalinZara said, divide both sides by $13^x$:
$$\left(\frac{10}{13}\right)^x+\left(\frac{11}{13}\right)^x+\left(\frac{12}{13}\right)^x=1+\left(\frac{14}{13}\right)^x$$
For $0 < a < 1$, $a^x$ is decreasing, so the left side is a decreasing function. For $a > 1$, $a^x$ is increasing, so the right side is an increasing function. A decreasing and increasing function can only intersect once, so there is only one solution.
I tried $x=1$ and that didn't work. I tried $x=2$ and it did. Therefore, the solution is $x=2$.
A: Hint: Divide by $13^x$ - the left hand side is decreasing and the right hand side is increasing; hence there is at most one solution. Since $x=2$ magically works, that is the unique solution.
A: Main idea: dividing by $12^x$ gives an equivalent equation (I chose $12$ because it is the central value).
Let us show that the function defined by: 
$$f(x):=\underbrace{(10/12)^x-(14/12)^x}_{g_1(x)}+1+\underbrace{(11/12)^x-(13/12)^x}_{g_2(x)}.$$
is strictly decreasing. In fact, it suffices to show that any function of the form:
$$h(x)=(1-a)^x-(1+a)^x \ \ \text{with} \ \ a>0$$
is strictly decreasing (indeed $g_1=h$ for $a=1/6$ and $g_2=h$ for $a=1/12$)
It is an almost immediate consequence of the fact that:
$$h'(x)=\underbrace{(1-a)^x}_{>0} \underbrace{ln(1-a)}_{<0} \ - \ \underbrace{(1+a)^x}_{>0} \underbrace{ln(1-a)}_{>0}$$
is negative.
As $x=2$ is a solution, it is thus the unique solution.
