Solving equation of the form $(x+y)\ln(x+y) = (x+z)\ln(x+z)$ for $x$ In the process of some research I am doing, I have come across the following equation:
$$
\frac{m_c+m_k+m_x}{\Delta t_x}\ln\left[\frac{v_{1x} - \frac{m_c+m_k+m_x}{\alpha k} -v_0}{v_{2x}-\frac{m_c+m_k+m_x}{\alpha k}-v_0}\right]
=
\frac{m_c+m_k+m_y}{\Delta t_y}\ln\left[\frac{v_{1y} - \frac{m_c+m_k+m_x}{\alpha k} -v_0}{v_{2y}-\frac{m_c+m_k+m_y}{\alpha k}-v_0}\right]
$$
I would like to solve this for parameter $m_c$. Since it's of the form
$$
(x+y)\ln(x+y) = (x+z)\ln(x+z)
$$
I can see how there will be no solution in terms of elementary functions. However, through reading, it seems that perhaps I could use Lambert's W function to help?
So far, I have tried to factor the left hand side for $m_c$. For example, it can be written like
$$
\frac{m_c}{\Delta t_x}\ln\left[v_{1x} - \frac{m_c}{\alpha k} - \frac{m_k + m_x}{\alpha k} - v_0\right] + \frac{m_k+m_x}{\Delta t_x}\ln\left[v_{1x} - \frac{m_c}{\alpha k} - \frac{m_k + m_x}{\alpha k} - v_0\right]
$$
which gets us to something of the form
$$x\ln(x+y) + \ln(x+y)
$$
Though I'm not sure if this is helpful.
Additional information
I should add that my ultimate goal in this is to develop a Taylor series of a couple of terms in the variable $v_{1x}$. Suppose that we could solve for $m_c$. Then we would have some solution like
$$m_c = F(v_{1x})$$
for some function $F$. I would like to know $\text{d}F(m_c)/\text{d}v_{1x}$, and $\text{d}^2F(m_c)/\text{d}v_{1x}^2$ - if this is somehow an "easier" problem to solve.
Further edit
I was wrong to assert that this is somehow equivalent to
$$
(x+y)\ln(x+y) = (x+z)\ln(x+z)
$$
As Piquito has shown, this is true iff $x = 0$ (when $y \neq z$). I have shown this with identities from the W function.
 A: HINT.-Because of $a\ln x=x^a$ and the injectivity of the function logarithm one has
$$(x+y)\ln(x+y) = (x+z)\ln(x+z)\iff(x+y)^{x+y}=(x+z)^{x+z}\qquad(*)$$
For each value of $x+y$ the number in the LHS in $(*)$ is the image of the real function $F(t)=t^t$ for $t=x+y$ and so is obviously for $x+z$ in the RHS.
The function $F(t)=t^t$ has a minimum at the point $t=\frac 1e$ for which the derivative $F'(t)=t^t(\ln t+1)=0$ and is one-to-one for $t\gt 1$. Besides each point in the interval $(F(\frac 1e),1]$ has two preimages in the interval $[0,1]$.
It follows that for all value of $x+y\gt 1$ there is equality in $(*)$ if and only if $x+y=x+z$ (i.e.$y=z$) and the only cases in which one can have equality in $(*)$ with $x+y\ne x+z$ are those in which $x+y$ and $x+z$ are the two preimages above. In other words there are two positive numbers $\epsilon_1$ and $\epsilon_2$ such that $\frac 1e-\epsilon_1$ and $\frac 1e+\epsilon_2$ belong to the interval $[0,1]$ and $$x+y=\frac 1e -\epsilon_1\\x+z=\frac 1e+\epsilon_2$$
This way you have $$(\frac 1e -\epsilon_1)\ln(\frac 1e -\epsilon_1) = (\frac 1e+\epsilon_2)\ln(\frac 1e+\epsilon_2)$$
