When is $\ln x = x^{\frac13}$?

Question

When is $\ln x = x^{\frac13}$?

Is there any way to find this solution that isn't just an estimate?

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Wolfram says it has to do with lambert functions? How do I get in to those?

Answer 1

Notice that \begin{align*} \ln x &= x^{1/3}\implies\\ x &= \exp(x^{1/3})\implies\\ x^{1/3} &= \exp(x^{1/3}/3)\implies\\ -\frac{x^{1/3}}{3}\exp\left(-\frac{x^{1/3}}{3}\right) &= -\frac{1}{3} \end{align*}

Since the Lambert $W$ function is defined as the inverse of $xe^x$, we find:

$$-\frac{x^{1/3}}{3}=W\left(-\frac13\right)$$

so then

$$x = -27\, W^3\left(-\frac13\right)$$

Note that this actually contains two solutions, using the different branches of the $W$ function.

Answer 2

I doubt you get anything nicer. Maple yielded two solutions:
$6.4056720789810570272$ and $93.354460835003657852$
I fed these into the https://isc.carma.newcastle.edu.au/advancedCalc
and it found nothing. I guess it remains for you to learn more about LambertW,
starting with, e.g., https://en.wikipedia.org/wiki/Lambert_W_function

Answer 3

If you do not want to use Lambert function and still look for the solution(s) of equation $$\log(x) = x^{\frac13}$$ consider that you look for the zero(s) of function $$f(x)=\log(x) - x^{\frac13}$$ $$f'(x)=\frac{1}{x}-\frac{1}{3 x^{2/3}}$$ $$f''(x)=\frac{2}{9 x^{5/3}}-\frac{1}{x^2} $$The first derivative cancels when $x=27$ and $f(27)=\log (27)-3\approx 0.295837$, $f''(27)=-\frac{1}{2187}$.

So, $x=27$ corresponds to a maximum and then the equation shows two roots. Graphing would show that the roots are "close" to $5$ and $100$. So, you have your starting points for Newton method. It should provide the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 5.000000000 \\ 1 & 6.169025112 \\ 2 & 6.398944686 \\ 3 & 6.405666633 \\ 4 & 6.405672079 \end{array} \right)$$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 100.0000000 \\ 1 & 93.34450012 \\ 2 & 93.35446083 \\ 3 & 93.35446084 \end{array} \right)$$ which are the solutions for ten significant figures.

Answer 4

Similar to Argon's answer, though the path is a bit different

If $\log(x)=x^{1/3}$, then $$ x^{-1/3}\log\left(x^{-1/3}\right)=-\frac13 $$ Thus, if we let $u=\log\left(x^{-1/3}\right)$, then $$ u\,e^u=-\frac13 $$ Therefore, $$ x=e^{-3u}=-27u^3=-27\operatorname{W}\!\left(-\frac13\right)^3 $$ There are two real branches of Lambert W for negative arguments. These give the two solutions $$ x=6.4056720789810570276 $$ and $$ x=93.354460835003657851 $$