Solving an equation with Lambert's W function? Or by any other means?

I am trying to solve the following equation for x in terms of $y$ and $c$ (with $x,y \in [0,1]$)

$$\begin{equation} \log\left(\frac{x}{1-x-y}\right) + \frac{x}{1-x-y} + \frac{y}{1-x-y} = c \end{equation}$$

I can solve this easier equation

$$\begin{equation} \log\left(\frac{x}{1-x-y}\right) + \frac{x}{1-x-y} = c \end{equation}$$

Let

$$\begin{equation} z = \frac{x}{1-x-y} \end{equation}$$

Then I can solve for $x$ using Lambert's W function

$$\begin{align} \log(z) + z &= c \notag \\ z &= \exp(c)\exp(-z) \notag \\ z \exp(z) &= \exp(c) \notag \\ z &= W(\exp(c)) \notag \\ x &= \frac{(1-y)W(\exp(c))}{1+W(\exp(c))} \notag \end{align}$$

Can anyone help me solve the harder equation? Is Lambert's W function helpful here?

Thanks!

Answer

$$\begin{align} \ln\left(\frac{x}{1-x-y}\right) +\frac{x}{1-x-y} + \frac{y}{1-x-y} &= c \tag{1}\label{1} \end{align}$$

$$\begin{align} \ln\left(\frac{x}{1-x-y}\right) &= c+\frac{1-x-y-1}{1-x-y} ,\\ \ln\left(\frac{1-y}{1-x-y}-1\right) &= c+1-\frac{1}{1-x-y} ,\\ \ln\left(\frac{1-y}{1-x-y}-1\right) &= c-\left(\frac{1}{1-x-y}-1\right) ,\\ \ln\left(\frac{1-y}{1-x-y}-1\right) &= c-\frac{1}{1-y}\left(\frac{1-y}{1-x-y}-(1-y)\right) ,\\ \ln\left(\frac{1-y}{1-x-y}-1\right) &= c-\frac{1}{1-y}\left(\frac{1-y}{1-x-y}-(1-y)\right) ,\\ \ln\left(\frac{1-y}{1-x-y}-1\right) &= c-\frac{y}{1-y}-\frac{1}{1-y}\left(\frac{1-y}{1-x-y}-1\right) . \end{align}$$

Let

$$\begin{align} \frac{1-y}{1-x-y}-1&=z ,\\ c-\frac{y}{1-y}&=u ,\\ -\frac{1}{1-y}&=v \end{align}$$

and we have an equation

$$\begin{align} \ln z&=u+vz , \end{align}$$

which has a standard solution for $z$ in terms Lambert W function

$$\begin{align} z&=-\frac{\operatorname{W}(-v\exp(u))}{v} ,\\ x&=(1-y)\left(1-\frac1{1+z}\right) . \end{align}$$