How to solve this equation for $y$? I have the following equation that I would like to solve for $y$:
$$\alpha=\sum_{i=1}^{n}\exp(-y\beta_i)$$
$\alpha$, $y$ and $\beta_i$ are all positive reals (the $\beta_i$'s are independent random numbers), and $n$ is a large positive integer. I'm unable to find a way to isolate $y$. Is this equation solvable?
Probability distribution of $\beta_i$:
$\beta_i$ is defined as $\beta_i=\sqrt{2\omega_i}$, where $\omega_i$ assumes a uniform probability distribution in some interval $[\omega_a,\omega_b]$, where $0\leq\omega_a<\omega_b$.
 A: Updated post
After publishing my original post the author of the OP has specified the PDF for the $\beta_i$, also hereby he has assumed that all $\beta_i$ have the same distribution (which was my main simplifying assumption).
Here I take up this definition and update my solution.
EDIT (18.06.19): the following through derivation yields the corrected result which is even simpler than the previous one.
The distribution function for $b$ is from the definition $w = \frac{1}{2}b^2$ and $f(w)$ flat between $w_a$ and $w_b$ given from $dw = b\; db$ after normalization
$$f(b) = \frac{2 b}{b_b^2-b_a^2} = \frac{b}{w_b-w_a}$$
Now we can do the integral $(2)$ which gives
$$\alpha = n \int_{b_a}^{b_b} f(b) e^{- y b} \, db \\
= \frac{2 n \left(e^{- y b_a } (y b_a+1)-e^{-y b_b} (y b_b +1)\right)}{y^2 \left(b_b^2-b_a^2\right)}\tag{4a}
$$
We have obtained a transcendental equation for $y(\alpha)$. This is best solved in the inverted form $(4a)$ for $\alpha (y)$.
In the limit of a very narrow distribution ($w_a \to w_b (=w)$) we get from $(4a)$
$$\alpha = n e^{-b_b  y}\tag{4b}
$$
which can be inverted to give the explicit solution
$$y = \frac{1}{b_b}\log(\frac{\alpha}{n})\tag{4c}$$
My original post
Making the simplifying assumptions that the $\beta_i$ have the same PDF $f(\beta)$ we can take the expectation value of the equation over all $\beta_i$ and get
$$E(\alpha) = \alpha = \sum_{i=1}^n E(\exp(- y \beta_i))= n E(\exp(- y \beta))\tag{1}$$
More explicitly we have for a continuous variable $\beta$
$$E(\exp(- y \beta))=\int_{0}^\infty f(\beta) \exp(- y \beta))^\,d\beta\tag{2}$$
Assuming for simplicity that $\beta$ is exponentially distributed, i.e.
$$f(\beta) = a \exp(- a \beta )\tag{3}$$
the integral can be done, and we obtain
$$\alpha = n \frac{a}{a + y}\tag{4}$$ 
which can be solved explicitly for $y$ to give 
$$y = a (\frac{n}{\alpha}-1)\tag{5}$$
