How the solve the equation $\frac{(a+b\ln(x))^2}{x}=c$ 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?
 A: 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:
$$
x = \left[\frac{2b}{\sqrt{c}}W\left(\mp\frac{\sqrt{c}}{2b}\exp\left(-\frac{a}{2b}\right)\right)\right]^2
$$
