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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

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    $\begingroup$ $x=0$ 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
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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

$$x=-a\left(\frac1b+\frac{1}{4c}W\left(-\frac{4c}{b}e^{-4c/b}\right)\right)$$

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  • $\begingroup$ 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
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    $\begingroup$ @mhennibenghorbal Pleased to hear. And good to know that you have such seemingly clairvoyant insights. ;-)) $\endgroup$ – Mark Viola Mar 4 '16 at 13:44
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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.

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