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I need to solve the equation $$\frac{(a+b\ln(x))^2}{x}=c$$ where $a$, $b$, and $c$ are given. It is known that $a$ and $b$ are fixed and satisfy some condition such that the left hand side is decreasing. So $x$ is uniquely determined by $c$ when $c$ is chosen in certain range.

A related problem is $$\frac{(a+b\ln(x))}{x}=c$$ for which the solution is $$x=-\frac{bW(-\frac{ce^{-a/b}}{b})}{c}$$ where $W(z)$ is the product log function.

Any hint about how to solve the first equation?

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    $\begingroup$ What about replacing $x$ by $y = \sqrt{x}$? $\endgroup$ Dec 14, 2021 at 13:18
  • $\begingroup$ Yes, it should work. Thx $\endgroup$
    – wyw
    Dec 14, 2021 at 13:58

1 Answer 1

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Step-by-step solution with Lambert W. The goal is to get something of the form $\color{red}{ue^u = v}$ then re-write it as $\color{blue}{u=W(v)}$.
$$ \frac{(a+b\ln(x))^2}{x}=c \\ \frac{(a+b\ln(x))}{\sqrt{x}}=\pm\sqrt{c} \\ (a+b\ln(x))e^{-\ln(x)/2}=\pm\sqrt{c} \\ (a+b\ln(x))\exp\left(-\frac{a}{2b}-\frac{\ln(x)}{2}\right) =\pm\sqrt{c}\exp\left(\frac{-a}{2b}\right) \\ (a+b\ln(x))\exp\left(-\frac{a+b\ln(x)}{2b}\right) =\pm\sqrt{c}\exp\left(\frac{-a}{2b}\right) \\ \color{red}{-\frac{a+b\ln(x)}{2b}\exp\left(-\frac{a+b\ln(x)}{2b}\right) =\mp\frac{\sqrt{c}}{2b}\exp\left(\frac{-a}{2b}\right)} \\ \color{blue}{-\frac{a+b\ln(x)}{2b} = W\left(\mp\frac{\sqrt{c}}{2b}\exp\left(\frac{-a}{2b}\right)\right)} \\ \ln(x) = \frac{-a-2bW\left(\mp\frac{\sqrt{c}}{2b}\exp\left(\frac{-a}{2b}\right)\right)}{b} \\ x = \exp\left(-\frac{a}{b}-2 W\left(\mp\frac{\sqrt{c}}{2b}\exp\left(\frac{-a}{2b}\right)\right)\right) $$


The other example mentioned...

$$ \frac{a+b\ln(t)}{t}=c \\ (a+b\ln(t))e^{-\ln(t)} = c \\ (a+b\ln(t))\exp\left(-\frac{a}{b}-\ln(t)\right) = c \exp\left(-\frac{a}{b}\right) \\ (a+b\ln(t))\exp\left(-\frac{a+b\ln(t)}{b}\right) = c \exp\left(-\frac{a}{b}\right) \\ \color{red}{-\frac{a+b\ln(t)}{b}\exp\left(-\frac{a+b\ln(t)}{b}\right) = -\frac{c}{b} \exp\left(-\frac{a}{b}\right)} \\ \color{blue}{-\frac{a+b\ln(t)}{b} = W\left(-\frac{c}{b} \exp\left(-\frac{a}{b}\right)\right)} \\ \ln(t) = -\frac{a+bW\left(-\frac{c}{b} \exp\left(-\frac{a}{b}\right)\right)}{b} \\ t = \exp\left(-\frac{a+bW\left(-\frac{c}{b} \exp\left(-\frac{a}{b}\right)\right)}{b}\right) $$

We may question the other solution given in the OP. In fact, this solution is equal to that solution:
Claim $$ \exp\left(-\frac{a+bW\left(-\frac{c}{b} \exp\left(-\frac{a}{b}\right)\right)}{b}\right) = -\frac{b}{c}W\left(-\frac{c}{b} \exp\left(-\frac{a}{b}\right)\right) \tag1$$ Why?
$$ \text{Let}\quad Q = W\left(-\frac{c}{b} \exp\left(-\frac{a}{b}\right)\right). \\ \text{Then}\quad Qe^Q = -\frac{c}{b}\exp\left(-\frac{a}{b}\right) \\ -\frac{b}{c}Q = \exp\left(-\frac{a}{b}\right)e^{-Q} \\ -\frac{b}{c}Q = \exp\left(-\frac{a+bQ}{b}\right) \\ \text{which is $(1)$.} $$


Challenge:
Simplity the first solution in the same way: $$ x = \left[\frac{2b}{\sqrt{c}}W\left(\mp\frac{\sqrt{c}}{2b}\exp\left(-\frac{a}{2b}\right)\right)\right]^2 $$

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  • $\begingroup$ Describing and color-coding your strategy like that is so helpful! I will definitely remember that trick. $\endgroup$ Dec 14, 2021 at 21:32
  • $\begingroup$ Thanks for the answer. Very nice and clear. But any idea about the solution of $\frac{a+b\ln(t)}{t}=c$ in the question? It seems a solution with different form will be obtained using your procedure. I obtain that solution from Wolfram Alpha. $\endgroup$
    – wyw
    Dec 15, 2021 at 2:30

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