Addition of different bases raised to the same power (solve for power). Is power determined uniquely? Let a vector $\mathbf{x} = [x_1, \  x_2, \ \dots \ , \  x_n]' \in \mathbb{R}^n$ contain non-negative elements i.e. zeros and positive values. 
Problem:  $x_1^a + x_2^a + \dots + x_n^a = c \quad$ where $c$ is known, the unknown $a$ is a scalar.
Can $a$ be determined uniquely? If yes, how may I prove it? Thanks in advance.
 A: I don't think so? 
Take $\vec{x} = [1/2,2] \in \mathbb{R}^2$.  Then:
$(\frac{1}{2})^a + 2^a = (\frac{1}{2})^{-a} + 2^{-a}$.
enedil's comment is wrong, as $f(a) = x^a$ is actually strictly decreasing for $x < 1$.  Are there any more restrictions on $a$?
A: Given
$$
0 < f(a) = x_{\,1} ^a  + x_{\,2} ^a  +  \ldots  + x_{\,n} ^a  = c\quad \left| {\,1 < x_{\,1}  \le  \cdots  \le x_{\,n} } \right.
$$
where the $x_k$ can be ordered in a non-decreasing way, without loss of generality,
then we have
$$
\eqalign{
  & f(a) = x_{\,1} ^a  + x_{\,2} ^a  +  \ldots  + x_{\,n} ^a  =   \cr 
  &  = x_{\,n} ^a \left( {1 + \left( {{{x_{\,n - 1} } \over {x_{\,n} }}} \right)^a  +  \ldots  + \left( {{{x_{\,1} } \over {x_{\,n} }}} \right)^a } \right) =   \cr 
  &  = n\,\xi ^{\,a} \quad \left| {\;x_{\,1}  \le \xi  \le x_{\,n} } \right. \cr} 
$$
and as commented, $f(a)$ is in this case increasing.
So we can write
$$
\eqalign{
  & n\,x_{\,1} ^a  < f(a) = c < n\,x_{\,n} ^a   \cr 
  & x_{\,1} ^a  < {c \over n} < x_{\,n} ^a   \cr 
  & a\ln \left( {x_{\,1} } \right) < \ln \left( {{c \over n}} \right) < a\ln \left( {x_{\,n} } \right)  \cr 
  & {{\ln c - \ln n} \over {\ln \left( {x_{\,n} } \right)}} < a < {{\ln c - \ln n} \over {\ln \left( {x_{\,1} } \right)}} \cr} 
$$
and use this estimate to start a Newton-Raphson approximation
to the (unique) zero of
$$
\ln f(a) - \ln c = 0
$$
