Simple Expression - Lambert W function

Question

I have all the below variables except for $a$, is there a closed form expression for $a$ with respect of the rest of the variables?

$$\lambda=\frac{1+(\ln (a)-\mu) / \sigma^{2}}{a}$$

What I have so far:

$\lambda a = 1 + (\ln(a) - \mu) / \sigma^2$

$\sigma^2[\lambda a-1]= \ln(a) - \mu$

$\sigma^2\lambda a-\sigma^2= \ln(a) - \mu$

$\sigma^2\lambda a- \ln(a)= \sigma^2 - \mu $

Thanks

Answer 1

\begin{align}\lambda=\frac{1+\frac{\ln a-\mu}{\sigma^2}}{a}&\implies\lambda=\frac{\sigma^2+\ln a-\mu}{\sigma^2a}\\&\implies\lambda\sigma^2a=(\sigma^2-\mu)+\ln a\\&\implies\frac1{e^{\sigma^2-\mu}}=ae^{-\lambda\sigma^2a}\\&\implies-\frac{\lambda\sigma^2}{e^{\sigma^2-\mu}}=-\lambda\sigma^2ae^{-\lambda\sigma^2a}\\&\implies W\left(-\frac{\lambda\sigma^2}{e^{\sigma^2-\mu}}\right)=-\lambda\sigma^2a\\&\implies a=-\frac1{\lambda\sigma^2}W\left(-\frac{\lambda\sigma^2}{e^{\sigma^2-\mu}}\right)\end{align}

Answer 2

As $a$ appears outside and inside a logarithm, there is no closed-form expression using the standard functions. This can probably be solved with $W$. https://en.wikipedia.org/wiki/Lambert_W_function