All real number for which $n$ in $5^n+7^n+11^n=6^n+8^n+9^n$ 
Finding all real  number $n$ in
$$5^n+7^n+11^n=6^n+8^n+9^n$$

Try: From given equation
$n=0,1$ are the solution
But i did not understand any other solution exists or not
Although i have tried like this way
$$\bigg(\frac{5}{9}\bigg)^n+\bigg(\frac{7}{9}\bigg)^n+\bigg(\frac{11}{9}\bigg)^n = \bigg(\frac{6}{9}\bigg)^n+\bigg(\frac{8}{9}\bigg)^n+1$$
Right side is strictly increasing function. but i have a confusion whether left side is strictly increasing or not
could some help me how to solve it, thanks

Consider the function $f(x)=x^n$ for positive $x$. Its second derivative is $n(n-1)x^{n-2}$ and therefore for $n > 1$ or $n<0$ $f$ is strictly convex while for $0 < n < 1$ f is strictly concave.
Our equation is equivalent to $f(5) + f(7) + f(11) = f(6) + f(8) + f(9)$
In the convex case we have:
$f(6) = f(\frac{5+7}{2}) < \frac{f(5)+f(7)}{2}$,
$f(9) = f(\frac{7+11}{2}) < \frac{f(7)+f(11)}{2}$, 
and that
$f(8) = f(\frac{11+5}{2}) < \frac{f(11)+f(5)}{2}$
So $ f(5) + f(7) + f(11) > f(6) + f(8) + f(9)$
In the concave case the same strict inequalities hold, but reversed.
Thus only n=0 or n=1 could be solutions and they both work.
