Optimization of $\left(1-\frac{x}{z}\right) \log_{2} \left(1+\frac{x}{m}\right)$ with respect to $x$ If I have the following form. I need to find the optimum value of x that maximise the function
$$ F(x)= \left(1-\frac{x}{z}\right) \log_{2} \left(1+\frac{x}{m}\right) $$
where $z,m >0$ are just a constant integer numbers and greater than zero. 
I have tried to derive this function with respect to $x$ and make it equal to zero. In fact, it is a bit complicated and I am not sure that there is a solution to solve it analytically or not. If not, any suggestion from where I have to start.   
Please, I need your help to solve this equation or to give me any hint to solve it. 
 A: The derivative of $F$ writes
\begin{equation}
F'(x) = -\frac{1}{z \ln2} \ln\left(1+\frac{x}{m}\right) + \frac{1}{z\ln2} \frac{z-x}{x+m} \, .
\end{equation}
Solving $F'(x)=0$ with respect to $x$ amounts to
\begin{aligned}
\ln\left(1+\frac{x}{m}\right) &  = \frac{z+m-(x+m)}{x+m} \\
& = \frac{z+m}{m}\left(1+\frac{x}{m}\right)^{-1} - 1 \, ,
\end{aligned}
i.e.
\begin{equation}
\left(\ln\left(1+\frac{x}{m}\right) + 1\right)\exp\left(\ln\left(1+\frac{x}{m}\right) + 1\right) = \frac{m+z}{m} \text{e} \, .
\end{equation}
This equation is of the form $X \text{e}^X=Y$, with $X=\ln\left(1+\frac{x}{m}\right) + 1$ and $Y=\frac{m+z}{m} \text{e}$. The solutions are
$X=W(Y)$, where $W$ denotes the Lambert W-function. Hence, a local extreme value of $F$ is reached at $x=x^*$, where
$$
x^* = m\exp\left(W\!\left(\frac{m+z}{m} \text{e}\right) - 1\right) - m
= \frac{m+z}{W\!\left(\frac{m+z}{m} \text{e}\right)} - m \, .
$$
Since
\begin{equation}
\lim_{x\rightarrow -m^{+}} F(x) = -\infty \qquad\text{and}\qquad \lim_{x\rightarrow +\infty} F(x) = -\infty \, ,
\end{equation}
the value $x=x^*$ corresponds to the maximum of $F$ over $\left]-m, +\infty\right[$. Finally,
$$ F(x^*) = \max_{x\in\left]-m, +\infty\right[} F(x) = \frac{m + z}{z\ln 2} \frac{\left(W\!\left(\frac{m+z}{m} \text{e}\right) - 1\right)^{2}}{W\!\left(\frac{m+z}{m} \text{e}\right)} \, . $$
