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!
 A: \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}
A: You can use the Lambert $W$ to solve
$$ \log(s) + s + t = c $$
and get a curve parameterized by $t$.
For each value of $(s,t)$, then,
$$ \frac{x}{1-x-y} = s \qquad\qquad \frac{y}{1-x-y} = t $$
now reduces to a system of linear equations in $x$ and $y$, which is easy to solve.
All in all, you get $(x,y)$ as afunction of $t$.
A: I fill in the details of Henning Makholm's answer here
First, the parametrized curve
\begin{align}
  \log(s) + s + t &= c \notag \\
  s \exp(s) &= \exp(c - t) \notag \\
  s &= W(\exp(c-t)) \notag
\end{align}
Second, the linear system gives
\begin{equation}
  x = \frac{s}{1+s+t}, \quad \quad  y = \frac{t}{1+s+t}
\end{equation}
Substituting $s$ gives
\begin{equation}
  x = \frac{W(\exp(c-t))}{1+W(\exp(c-t))+t}, \quad \quad  y = \frac{t}{1+W(\exp(c-t))+t}
\end{equation}
This solves for $x$ and $y$ in terms of $c$ and $t$. The solution is implicit in the sense that $t = \frac{y}{1-x-y}$.
