I have following equation to solve for $x$ $$\ln\left(1+\frac{bx}{a}\right)=\frac{4cx}{a}$$ where $a>0,b>0$ and $c>0$. In my own attempt I replaced $1+\frac{bx}{a}$ by $y$ and with this replacement the final form of the equation is $$ye^{-\frac{4cy}{b}}=e^{-\frac{4c}{b}}$$ I don't know how to proceed further. Any help in this regard will be much appreciated.
BR
Frank
You are almost there! Here is how you proceed. Let $z=-\frac{4by}{c}$ and simplify to get
$$ze^z=f(b, c) \implies z=W(f(b, c)) $$
I think you can finish it. See here.
We start with the equation of interest
$$\log\left(1+\frac{bx}{a}\right)=\frac{4cx}{a} \tag 1$$
Now, let $\alpha = b/a$ and $\beta = 4c/a$. Then, we can rewrite $(1)$ as
$$1+\alpha x=e^{\beta x} \tag 2$$
Multiplying $(2)$ by $\frac{\beta}{\alpha}e^{\beta/\alpha}$ yields
$$\left(\frac{\beta}{\alpha}+\beta x\right)e^{\beta/\alpha}=\frac{\beta}{\alpha}e^{\left(\frac{\beta}{\alpha}+\beta x\right)}\tag 3$$
Rearranging $(3)$ we obtain
$$-\left(\frac{\beta}{\alpha}+\beta x\right)e^{-\left(\frac{\beta}{\alpha}+\beta x\right)}=-\frac{\beta}{\alpha}e^{-\beta/\alpha} \tag 4$$
Invoking the definition of Lambert's $W$ reveals
$$-\left(\frac{\beta}{\alpha}+\beta x\right)=W\left(-\frac{\beta}{\alpha}e^{-\beta/\alpha}\right)$$
whereupon solving $(5)$ for $x$, we obtain
$$x=-a\left(\frac1b+\frac{1}{4c}W\left(-\frac{4c}{b}e^{-4c/b}\right)\right)$$
