Bailee Short

2022-06-28

Solving equations of type ${x}^{1/n}={\mathrm{log}}_{n}x$
First, I'm a new person on this site, so please correct me if I'm asking the question in a wrong way.
I thought I'm not a big fan of maths, but recently I've stumbled upon one interesting fact, which I'm trying to find an explanation for. I've noticed that graphs of functions $y={x}^{1/n}$ and $y={\mathrm{log}}_{n}x$ , where $n$ is given and equal for both functions, always have $2$ intersection points. This means, equation ${x}^{1/n}={\mathrm{log}}_{n}x$ must have $2$ solutions, at least it's what I see from the graphs.
I've tried to solve this equation analytically for some given $n$, like $4$, but my skills are very rusty, and I cannot come up with anything. So I'm here for help, and my question(-s) are:
are these $2$ functions always have $2$ intersection points?
if yes, why, if not, when not?
how to solve equations like ${x}^{1/n}={\mathrm{log}}_{n}x$ analytically?

benedictazk

$\left(-u\right)\cdot {e}^{-u}=-\frac{\mathrm{ln}n}{n}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}u=-W\left(-\frac{\mathrm{ln}n}{n}\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x={t}^{n}=\left({e}^{u}{\right)}^{n}={e}^{nu}$
$x=\mathrm{exp}\left(-n\cdot W\left(-\frac{\mathrm{ln}n}{n}\right)\right)$
where W is the Lambert W function.

abbracciopj

They always have two intersection points. Let
$f\left(x\right)={x}^{1/n}-{\mathrm{log}}_{n}x.$
Then
$\underset{x\to {0}^{+}}{lim}f\left(x\right)=\underset{x\to +\mathrm{\infty }}{lim}f\left(x\right)=+\mathrm{\infty }.$
Also,
${f}^{\prime }\left(x\right)=\frac{1}{n}\phantom{\rule{thinmathspace}{0ex}}{x}^{1/n-1}-\frac{1}{x\phantom{\rule{thinmathspace}{0ex}}\mathrm{log}n}.$
It is easy to see that ${f}^{\prime }$ has a single zero, which is necessarily a minimum (one can check that ${f}^{\prime }$ has the appropriate signs at both sides of this point, which is
${x}_{m}=\frac{{n}^{n}}{\left(\mathrm{log}n{\right)}^{n}}.$
We have
$f\left({x}_{m}\right)=\frac{n}{\mathrm{log}n}-\frac{n}{\mathrm{log}n}\phantom{\rule{thinmathspace}{0ex}}\left(\mathrm{log}n-\mathrm{log}\mathrm{log}n\right)<0,$
so the minimum is achieved below the $x$-axis. This shows that $f$ intersects the $x$-axis twice.
As for an analytic solution, I don't think that's possible.

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