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.



  • 2
    $\begingroup$ $x=0$ is a solution. $\endgroup$ – Thomas Andrews Mar 4 '16 at 3:20
  • $\begingroup$ @ThomasAndrews any other solution other than zero. $\endgroup$ – Frank Moses Mar 4 '16 at 3:22
  • $\begingroup$ can we use Lambert function to find the solution $\endgroup$ – Frank Moses Mar 4 '16 at 3:22
  • $\begingroup$ @ThomasAndrews how about if we multiply both sides by $-\frac{4c}{b}$ because then, I think, equation can be written in the form of $xe^x$ $\endgroup$ – Frank Moses Mar 4 '16 at 3:26
  • $\begingroup$ Yes, you can use one of the branches to find another solution when $\frac{4c}{b}>0$. Depends on whether $\frac{4c}{b}>1$ or $\frac{4c}{b}<1$ whether you use the $W_{0}$ or $W_{-1}$, respectively. $\endgroup$ – Thomas Andrews Mar 4 '16 at 3:27

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


  • $\begingroup$ Th OP is doing fine! He is just stuck at some point which has been clarified for him and he can continue now! $\endgroup$ – Mhenni Benghorbal Mar 4 '16 at 8:00
  • 4
    $\begingroup$ @mhennibenghorbal Pleased to hear. And good to know that you have such seemingly clairvoyant insights. ;-)) $\endgroup$ – Mark Viola Mar 4 '16 at 13:44

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.