Addition of different bases raised to the same power (solve for power). Is power determined uniquely?

Question

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.

Answer 1

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 $$ \ln f(a) - \ln c = 0 $$

Answer 2

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 $a$?