Is $\sum_{i=1}^n [-1-x_i e^{-\alpha}]=0$ solvable analytically or numerically for $\alpha$?

Question

Is $\sum_{i=1}^n [-1-x_i e^{-\alpha}]=0$ solvable analytically or numerically for $\alpha$?

My own interpretation is that if one tries to solve the above by taking the logarithm, then one is not able to get rid of the logarithm, but rather it has to be approximated numerically.

Is this correct?

Answer 1

Its actually totally straight-forward $$\begin{align} 0 &= \sum_{i=1}^n(-1-x_ie^{-\alpha}) \\ &=-n-e^{-\alpha}\sum_{i=1}^nx_i \end{align}$$ This can be explicitly solved for alpha: $$ \alpha = \log\left(\frac{-1}{n}\sum_{i=1}^nx_i\right)$$