Manipulating problem for Lambert w function (Finding roots of equation)

Question

In my calculus book, a problem is to graph $y = x^3/24-\ln x$. One part of it was to find the roots of the equation. I was unable to do anything to solve this problem, and wolfram alpha used the Lambert w function. I searched it up, and what I could understand was that it's the inverse of $f(x) = xe^x$. However, I don't understand how I can manipulate this equation to get an answer. I was unable to find anything that had an answer. How can I solve this?

Answer 1

$$ y=x^3/24-\log\left(x\right)\tag1 $$ $$ -\frac18e^{-3y}=-\frac{x^3}8e^{-x^3/8}\tag2 $$ $$ \operatorname{W}\!\left(-\frac18e^{-3y}\right)=-\frac{x^3}8\tag3 $$ $$ \bbox[5px,border:2px solid #C0A000]{-2\operatorname{W}\!\left(-\frac18e^{-3y}\right)^{\!1/3}=x}\tag4 $$ Explanation:
$(2)$: multiply both sides by $-3$, exponentiate, multiply both sides by $-\frac18$.
$(3)$: apply LambertW to both sides
$(4)$: take the cube root and multiply by $-2$

Note that for negative arguments, there are two branches of LambertW. For example, for $y=0$, the two branches give $$ x=1.0493180227889868869 $$ and $$ x=2.9660424669277519890 $$ which I computed in Mathematica as

N[2(-LambertW[0,-1/8])^(1/3),20]

and

N[2(-LambertW[-1,-1/8])^(1/3),20]

Answer 2

I think the Lambert W Function might be a outside the scope of most calculus courses, although admittedly I hadn't heard of it before seeing this post so I don't know for sure. One thing you could try is using Newton's method or your calculator's graphing utility to approximate the zeros of $\frac{x^{3}}{24} - \ln(x)$. I doubt that there is a way to find exact expressions for the solutions to equations like that, at least not with techniques that I'm familiar with.