Where does $\frac{x}{\log(x)}$ overtake $x^p$ for $0<p<1$?

Question

I am teaching a mathematics for CS class, and one of the questions I put on the homework that nearly every student got wrong was if asymptotically the function $x^{.999}$ was less than $\frac{x}{\log(x)}$ (prime counting function).

I thought this problem was simple as the limit is quite easy $$ \lim_{x\rightarrow\infty} \frac{x^{.999}}{\frac{x}{\log(x)}} = 0. $$

However, I realized that many students simply plotted this function (rather than taking the limit like I told them to) and found that for very large values of $x$, that $x^{.999}$ is bigger, therefore it must be bigger.

I wanted to solve out to see exactly where these two curves intersect to provide a cautionary tale into why you should do the math rather than taking the easy heuristic, however I had trouble solving for where this point was, even numerically as it grows quite large.

Suppose that $0<p<1$, we know that asymptotically that $x^p < \frac{x}{\log(x)}$ so performing some algebra can we find the point

$$ \log(x) = x^{1-p} $$

or in a form that looks almost like a Lambert W function

$$ 1 = \frac{1}{x}e^{x^{1-p}}. $$

A large lower bound as a function of $p$ could also be appreciated!

Answer 1

The curves intersect when $$\log \log x =(1-p) \log x $$ Let $t=\log x$. You want $$\log t = (1-p) t \\ t= e^{(1-p)t} \\ (p-1)te^{(p-1)t} =p-1 \\ (p-1)t = W(p-1) \\ t=\frac{W(p-1)}{p-1}$$ and so $$ x= \exp \left( \frac{W(p-1)}{p-1} \right) $$

Answer 2

Take $x^p = \frac{x}{\log x}$, that is $x^{p-1}\log x = 1$. Multiply by $p-1$ to get $x^{p-1}\log x^{p-1} = p-1$ and write it as $\log x^{p-1} e^{\log x^{p-1}} = p-1$.

Apply the Lambert function $W_{-1}$ to get $\log x^{p-1} = W_{-1}(p-1)$, then $\log x = \frac{1}{p-1}W_{-1}(p-1)$ and finally $x = e^{\frac{1}{p-1}W_{-1}(p-1)}$.

For $p=.999$ we get $x \approx 7.94 \times 10^{3959}$