I'm having a bit of trouble computing the inverse of $y = \mathrm{e}^{(2\mu+\sigma^2)}(\mathrm{e}^{(\sigma^2)}-1)$. Here's what I've done so far: \begin{align*} y &= \mathrm{e}^{(2\mu+\sigma^2)}(\mathrm{e}^{(\sigma^2)}-1) \\ \sigma^2 &= \mathrm{e}^{(2\mu+y)}(\mathrm{e}^{(y)}-1) \\ \ln{(\sigma^2)} &= (2\mu+y)+ \ln{(\mathrm{e}^{(y)}-1)} \\ \ln{(\sigma^2)} - 2\mu &= y + \ln{(\mathrm{e}^{(y)}-1)} \end{align*}
And from there I'm stuck :( can anyone offer help?
Swapping $\sigma^2$ and $y$ in the original equation does not make sense. Just keep them where they are and solve for $\sigma^2$. After all, you want to know which value $\sigma^2$ has to attain to get a specific $y.$ \begin{eqnarray} y & = & e^{(2\mu+\sigma^2)}\big(e^{\sigma^2}-1\big) \\ e^{-2\mu}y & = & e^{\sigma^2}\big(e^{\sigma^2}-1\big) \\ \left(e^{\sigma^2}\right)^2 - e^{\sigma^2} - e^{-2\mu}y & = & 0 \end{eqnarray} Solving the quadratic equation: $$ e^{\sigma^2} = \frac{1}{2}\pm\sqrt{\frac{1}{4}+e^{-2\mu}y} $$ As we are going to use the logarithm, the "minus" makes no sense: $$ e^{\sigma^2} = \frac{1}{2} + \sqrt{\frac{1}{4}+e^{-2\mu}y} $$ or $$ \sigma^2 = \ln \left( \frac{1}{2} + \sqrt{\frac{1}{4}+e^{-2\mu}y} \right) $$
