Finding the closed-form expression of x

Question

I am struggling to find a closed-form expression for $x$ in the following equation. I feel that the Lambert W function may be useful for finding an answer. I have been working on this for hours; however, I could not find any solution. I would really appreciate it if you could help me solve this. Thanks. $$\frac{ax+b}{(cx+d)(kx+d)}-\ln\left(\frac{kx+d}{cx+d}\right)=y$$

Answer 1

While I don't want to bash out an exact solution, I think I have an idea to help you solve the problem.

We can rewrite your equation as $\dots $

$$ \frac{ax+b}{(cx+d)(kx+d)} - \left(ln(kx+d) - ln(cx+d)\right)= y $$

Let $f(x) = ln(kx+d) - ln(cx+d)$. Since $\frac{d}{dx} (ln(x)) = \frac{1}{x}$, we can write

$$f'(x) = \frac{d}{dx} f(x) = \frac{k}{kx+d} - \frac{c}{cx+d} = \frac{d(k -c)}{(kx+d)(cx+d)}.$$

Therefore if we have $a = 0$ and $b = d(k-c)$, we have an equation of the form $\dots$

$$f'(x) - f(x) =y $$

which has solutions of the form (https://www.wolframalpha.com/input/?i=f%27(x)+-+f(x)+%3D+y)

$$y = c_1 e^x - y$$

Hope this helps find some solutions!

Answer 2

Assuming that all constants are non $0$, $c\neq k$, considering $$y=\frac{ax+b}{(cx+d)(kx+d)}-\ln\left(\frac{kx+d}{cx+d}\right)$$ let $$t=\frac{kx+d}{cx+d}\implies x=\frac{d(1- t)}{c t-k}$$ making $$y=\frac{\left(b k^2-a d k\right)+ (a c d+a d k-2 b c k)\,t+ \left(b c^2-a c d\right)\,t^2 }{d^2 (c-k)^2 \,t}-\log(t)$$ The roots of the numerator are $$t_1=\frac kc\qquad \text{and} \qquad t_2=\frac{b k-a d}{b c-a d}$$ which means that $y$ write in the form of

$$y=\frac{\alpha(t-t_1)(t-t_2)}t-\log(t)$$ which means that $t$ "can" be expressed in terms of the generalized Lambert function.