Solving or approximating an equation

Question

I was wondering: is it possible to solve for $y$ this equation?

$$ x = (y - {ab \over c} \log (1 - {dc \over f} y)){1 \over k} $$

I would like to graph $y(x)$ from this equation, but I am not able to solve it explicitly. Is there a method, even in non elementary functions? (I was thinking about Lambert W function for instance).

If it's not possible, could I for example approximate the solution? How could I apply a Taylor Expansion to make the equation a little bit easier?

To create a graph, I worked on Desmos by inserting the equation and then graphing $y = f(x)$, but I was wondering if I could do a little more work on it before giving up.

Thanks in advance, Dave

Answer 1

First get rid of the messy constants. By setting $z:=1-\dfrac{dc}fy$, you can reduce to a form

$$\log(z)=pz+q$$

or

$$-pze^{-pz}=-pe^q,$$

$$-pz=W(-pe^q).$$

Answer 2

You are correct about Lambert function.

Making, for more legibility, $\alpha=\frac{a b }{c}$ and $\beta=\frac{c d }{f}$, the solution of $$y-\alpha\log \left(1-\beta y\right)=k x$$ is given by $$y=\frac{1}{\beta }-\alpha W\left(\frac{e^t}{\alpha \beta }\right)\qquad \text{where} \qquad t=\frac{1}{\alpha \beta }-\frac{k }{\alpha }x$$

Answer 3

\begin{eqnarray} -kx &=& -y + \frac{ab}{c}\ln\left[1 - \frac{cdy}{f} \right] \\ -\frac{ckx}{ab} &=& -\frac{cy}{ab} + \ln\left[1 - \frac{cdy}{f} \right] \\ \ln f -\frac{ckx}{ab} &=& \ln(f - cdy) - \frac{cy}{ab} \\ f e^{-ckx/ab}&=& (f - cdy) e^{-cy/ab} \\ f e^{f/abd-ckx/ab} &=& (f - cdy) e^{f/abd-cy/ab} \\ \frac{f e^{f/abd-ckx/ab}}{abd} &=& \frac{(f - cdy)}{abd} e^{f/abd-cy/ab} \\ \color{red}{\frac{f e^{f/abd-ckx/ab}}{abd}} &=& \color{blue}{\frac{(f - cdy)}{abd}} e^{\color{blue}{(f - cdy)/abd}} \\ \color{blue}{\frac{f - cdy}{abd}} &=& W\left(\color{red}{\frac{f e^{f/abd-ckx/ab}}{abd}}\right) \\ y &=& \frac{1}{cd}\left[f - abd W\left(\frac{f e^{f/abd-ckx/ab}}{abd} \right)\right] \end{eqnarray}