I have the following equation:
$$\ln \left(\frac{\alpha^2}{\beta^2}\right) = \frac{1}{m \xi_\alpha} + \frac{b\ln(m \xi_\alpha)}{m^2}$$
where $\xi_\alpha := \xi(\alpha^2)$. The objective is to solve for $\xi_\alpha$ given that $\ln (\frac{\alpha^2}{\beta^2}) \gg 1$.
Here is my attempt:
$$ \ln\left(\frac{\alpha^2}{\beta^2}\right) = \frac{1}{m \xi_\alpha} + \frac{b\ln(m \xi_\alpha)}{m^2}$$
Exponentiating both sides, we get,
$$\frac{\alpha^2}{\beta^2} = e^\frac{1}{m\xi_\alpha} (m\xi_\alpha)^\frac{b}{m^2}$$
At this point, I am completely stuck. I am not sure how to use the fact that $\ln (\frac{\alpha^2}{\beta^2}) \gg 1$. I tried reading some materials online and there were some suggestions using iteratively solving the equation, but I am not sure how to proceed. Could someone help me with this?
Why do you want an iterative method since there is an explicit solution ?
Let $$a =\log\left(\frac{\alpha^2}{\beta^2}\right) \qquad \text{and} \qquad y=m \xi_\alpha $$ to make the equation $$a=\frac 1 y+\frac b{m^2} \log(y)$$ the solution of which being $$y=-\frac {m ^2}{b\,W(t)}\qquad \text{where} \qquad t=-\frac{m^2}{b}e^{-\frac{a m^2}{b}}$$ where $W(t)$ is Lambert function.
To make even simpler, let $k=-\frac {m^2}b$ which makes $$y=\frac{k}{W\left(k e^{a k}\right)}$$
Now, if $a$ is large and then $t$ is small, you can use the expansion of $W(t)$ around $t=0$
$$W(t)=t-t^2+\frac{3 }{2}t^3-\frac{8 }{3}t^4+O\left(t^5\right)$$