So today I came across an answer on MathStackExchange. I read it and found it well explained. But I stuck on a step where the user write "Solve using the properties of $W$ function". I don't know what property the user used.
What I know about Lambert $W$ function is that, $W(x) = f^{-1} (x)$, where $f(x) = x\cdot e^x$.
Can anyone solve the question or just tell me what the property the answerer used in the question which I have linked to you?
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as @Randall mentioned, using the $W$ function is really just putting a name on it, only more complex maths programs are actually going to return values for you. That said: $$y^x=x^y\\ y^{1/y}=x^{1/x}\\ \frac1y\ln y=\frac1x\ln x$$ now say $y=e^{-u}$ so: $$ue^u=-\frac1x\ln x\\ u=W_k\left(-\frac1x\ln x\right)\\ y=\exp\left[-W_k\left(-\frac1x\ln x\right)\right]$$ where $k$ denotes the branch of the $W$ function
The step in question is how do the properties of the Lambert $W$ function, give us the following $$-\dfrac{y\log(x)}{x}\exp\left(-\dfrac{y\log(x)}{x}\right) = -\dfrac{\log(x)}{x}\implies-xy\log(x) = \operatorname{W}\left(-\dfrac{\log(x)}{x}\right).$$
There is actually a typo in the linked answer, (which I corrected), the desired result should be $$-\dfrac{y\log(x)}{x}\exp\left(-\dfrac{y\log(x)}{x}\right)= -\frac{\log(x)}{x}\implies= -\frac{y\log(x)}{x} = \operatorname{W}\left(-\dfrac{\log(x)}{x}\right).$$
Here we will use the fact that $$z = we^w \iff w = W(z).$$
We first make the substitutions $$s = -\frac{y\log(x)}{x},\qquad z = -\frac{\log(x)}{x},$$
and rewrite the first equation as $se^{s} =z.$
Using the noted property of $W$, then, we can rewrite this as $s = W(z)$, which reversing our substitutions gives:
$$-\frac{y\log(x)}{x} = W\left(-\frac{\log(x)}{x}\right).$$