Is there a way to simplify the following? $$ \lambda^0 \theta=\beta \frac{e^{(\lambda^1-\lambda^0)\theta}}{1+e^{(\lambda^1-\lambda^0)\theta}} $$
I am particularly trying to understand the relationship between $\lambda^0$ and $\lambda^1$ and I can't seem to find a way to connect them more usefully.
$\lambda^0 \frac{1}{\beta}\theta-1= \frac{e^{(\lambda^1-\lambda^0)\theta}}{1+e^{(\lambda^1-\lambda^0)\theta}}-1=\frac{-1}{1+e^{(\lambda^1-\lambda^0)\theta}}$
$1+e^{(\lambda^1-\lambda^0)\theta}=\frac{1}{1-\lambda^0 \frac{1}{\beta}\theta}$
$e^{(\lambda^1-\lambda^0)\theta}=\frac{1}{1-\lambda^0 \frac{1}{\beta}\theta}-1=\frac{\lambda^0 \frac{1}{\beta}\theta}{1-\lambda^0\frac{1}{\beta} \theta}$
Take natural log
$(\lambda^1-\lambda^0)\theta=ln(\frac{\lambda^0 \frac{1}{\beta}\theta}{1-\lambda^0 \frac{1}{\beta}\theta})$
Multiply both the numerator and denominator by $\beta$ inside the logarithm.
$\lambda^1-\lambda^0=\frac{1}{\theta}ln(\frac{\lambda^0 \theta}{\beta-\lambda^0 \theta})$
This is a direct relation between $\lambda^1$ and $\lambda^0$.