Minimum for this function I thought of writing this question Minimum for this function in a different way, if it helps.

I want to minimize 
 $$\sum_{i=1}^n a_ix_i  + \nu \sum_{i=1}^n b_i 2^{x_i} ,$$
where $a_i \in [0,1]$, $b_i \in (0,\infty)$, and $x_i \in [x_{\min},0)$, and
$$ \sum_{i=1}^n 2^{x_i} = 1 .$$

$x_{\min}$, $\nu$, $a_i$  and  $b_i$ are constants.

I guess the tricky part is to minimize the function while ensuring all $2^{x_i}$ sum up to $1$ for these $n$ variables.

Thank you for your patience and help.
First let $x = (x_1, \dots, x_n)$ and write $f(x) = \sum_i^n a_i x_i + \nu\sum_i^n b_i 2^{x_i}$ and $g(x) = \sum_i^n 2^{x_i} - 1$. Now the method of Lagrange's multipliers tells you that you have a solution $x'$ iff it is an extremal point of the function $f(x) + \lambda g(x)$ with $\lambda$ a parameter to be determined. So you get equations $a_i + (\nu b_i + \lambda) 2^{x'_i} \log{2} = 0$ for all $1 \leq i \leq n$. It's now easy to express $x'_i$ as a function of lambda and plugging this into the remaining constraint $g(x') = 0$ to find the lambda and complete the solution. Hope this helps.
I don't think rewriting this way helps. In the first place, how would you show that if $y$ is the optimal solution of $\min_y f(y)$, then $x = 2^y$ is the optimal solution to $\min_x f(\log x)$?