Express $y$ from $\ln(x)+3\ln(y) = y$

Question

i am finding a inverse function to $$y=\frac{e^\frac{x^3}{3}}{x^3}$$ First step: $$x=\frac{e^\frac{y^3}{3}}{y^3}$$ Next: ... $$(xy^3)^3 = e^{{y}^3}$$

and now in this step i have no idea how can i express "y"

$$\ln(x)+3\ln(y) = y$$

Answer 1

Consider the function $$ f(x)=\frac{e^{x^3/3}}{x^3} $$ This function is defined for $x\ne0$. Also $$ \lim_{x\to-\infty}f(x)=0,\quad \lim_{x\to0^-}f(x)=-\infty,\quad \lim_{x\to0^+}f(x)=\infty,\quad \lim_{x\to\infty}f(x)=\infty $$ so the function has no “global” inverse.

Its derivative is $$ f'(x)=\frac{x^5e^{x^3/3}-3x^2e^{x^3/3}}{x^6}=\frac{x^3-3}{x^4}e^{x^3/3} $$ which vanishes for $x=\sqrt[3]{3}$ and is negative for $x<\sqrt[3]{3}$ (but $x\ne0$, of course) and positive for $x>\sqrt[3]{3}$.

So the function is decreasing in the intervals $(-\infty,0)$ and $(0,\sqrt[3]{3}]$; it is increasing in the interval $[\sqrt[3]{3},\infty)$.

How can we find the range? We know that the interval $(-\infty,0)$ is contained in the range (consider the $(-\infty,0)$ part of the domain). Since $$ f(\sqrt[3]{3})=\frac{e}{3} $$ we know that the range corresponding to the part $(0,\infty)$ of the domain is $[e/3,\infty)$.

Putting things together, the range is $$ (-\infty,0)\cup[e/3,\infty) $$

The restriction of $f$ to each of those intervals has an inverse function. Which is, well, the inverse function: there's no “explicit” form for it. But it's not necessary to find an inverse function, in order to determine the range.

Answer 2

Go back to the original:

$$y = x^{-3}e^{x^3/3}$$

Then: $$\frac{-1}{y}=-x^3e^{-x^3/3}$$ So: $$\frac{-1}{3y} = \frac{-x^3}{3}e^{-x^3/3}$$ This means that:

$$\frac{-x^3}{3} = W\left(\frac{-1}{3y}\right)$$

where $W$ is Lambert's $W$-function, with $W$ an inverse of $ze^z$.

So $$x=\left(-3W\left(\frac{-1}{3y}\right)\right)^{1/3}$$

You are not going to be able to do it without Lambert or some other non-elementary function.