I would like your help to understand whether the following system admits a unique solution for the unknowns $\rho_1, \rho_2, \rho_0$.
$$ \begin{cases} 1-t=\frac{\exp \{ \rho_0+\rho_1+ \rho_2(1)\}}{1+\exp \{ \rho_0+\rho_1+ \rho_2\}}\\ r=\frac{\exp \{ \rho_0+ \rho_2 \}}{1+\exp\{ \rho_0+ \rho_2\}} \\ w=\frac{1}{1+\exp \{ \rho_0+\rho_1+ 2\rho_2\}} \end{cases} $$ $r,t,w$ are known values in $(0,1)$.
Here my incomplete attempt to simplify:
$$ \begin{cases} \log(1-t)=\rho_0+\rho_1+ \rho_2-\log(1+\exp \{ \rho_0+\rho_1+ \rho_2\})\\ \log(r)= \rho_0+ \rho_2 -\log(1+\exp\{ \rho_0+ \rho_2\}) \\ \log(w)=-\log(1+\exp \{ \rho_0+\rho_1+ 2\rho_2\}) \end{cases} $$ I'm unable to proceed.
$\displaystyle r=\frac{\exp \{ \rho_0+ \rho_2 \}}{1+\exp\{ \rho_0+ \rho_2\}}\implies\exp\{ \rho_0+ \rho_2\}=\frac r{1-r}$
$\displaystyle1-t=\frac{\exp \{ \rho_0+\rho_1+ \rho_2\}}{1+\exp \{ \rho_0+\rho_1+ \rho_2\}}=\frac{r\exp{\{\rho_1\}}}{1-r+r\exp \{\rho_1\}}\implies\exp{\{\rho_1\}}=\frac{(1-r)(1-t)}{rt}$
$\displaystyle w=\frac{1}{1+\exp \{ \rho_0+\rho_1+ 2\rho_2\}}=\frac{1}{1+\exp \{ \rho_2\}\cdot\frac r{1-r}\cdot\frac{(1-r)(1-t)}{rt}}\implies \exp{\{\rho_2\}}=\frac{t(1-w)}{w(1-t)}$
$$\therefore\displaystyle\rho_0=\ln\Big[\frac{rw(1-t)}{t(1-r)(1-w)}\Big]$$
$$\displaystyle\rho_1=\ln\Big[\frac{(1-r)(1-t)}{rt}\Big]$$
$$\displaystyle\rho_2=\ln\Big[\frac{t(1-w)}{w(1-t)}\Big]$$