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?

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

Beginner2022-06-29Added 22 answers

$\begin{array}{rl}\sqrt[n]{x}& ={\mathrm{log}}_{n}x\\ \\ x& ={t}^{n}\end{array}\text{}{\textstyle \}}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}t=\frac{n}{\mathrm{ln}n}\cdot \mathrm{ln}t\phantom{\rule{1em}{0ex}};\phantom{\rule{1em}{0ex}}t={e}^{u}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{e}^{u}=\frac{n}{\mathrm{ln}n}\cdot u\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$

$(-u)\cdot {e}^{-u}=-\frac{\mathrm{ln}n}{n}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}u=-W{\textstyle (}-\frac{\mathrm{ln}n}{n}{\textstyle )}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x={t}^{n}=({e}^{u}{)}^{n}={e}^{nu}$

$x=\mathrm{exp}{\textstyle (}-n\cdot W{\textstyle (}-\frac{\mathrm{ln}n}{n}{\textstyle )}{\textstyle )}$

where W is the Lambert W function.

$(-u)\cdot {e}^{-u}=-\frac{\mathrm{ln}n}{n}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}u=-W{\textstyle (}-\frac{\mathrm{ln}n}{n}{\textstyle )}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x={t}^{n}=({e}^{u}{)}^{n}={e}^{nu}$

$x=\mathrm{exp}{\textstyle (}-n\cdot W{\textstyle (}-\frac{\mathrm{ln}n}{n}{\textstyle )}{\textstyle )}$

where W is the Lambert W function.

abbracciopj

Beginner2022-06-30Added 4 answers

They always have two intersection points. Let

$f(x)={x}^{1/n}-{\mathrm{log}}_{n}x.$

Then

$\underset{x\to {0}^{+}}{lim}f(x)=\underset{x\to +\mathrm{\infty}}{lim}f(x)=+\mathrm{\infty}.$

Also,

${f}^{\prime}(x)=\frac{1}{n}\phantom{\rule{thinmathspace}{0ex}}{x}^{1/n-1}-\frac{1}{x\phantom{\rule{thinmathspace}{0ex}}\mathrm{log}n}.$

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

${x}_{m}=\frac{{n}^{n}}{(\mathrm{log}n{)}^{n}}.$

We have

$f({x}_{m})=\frac{n}{\mathrm{log}n}-\frac{n}{\mathrm{log}n}\phantom{\rule{thinmathspace}{0ex}}(\mathrm{log}n-\mathrm{log}\mathrm{log}n)<0,$

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.

$f(x)={x}^{1/n}-{\mathrm{log}}_{n}x.$

Then

$\underset{x\to {0}^{+}}{lim}f(x)=\underset{x\to +\mathrm{\infty}}{lim}f(x)=+\mathrm{\infty}.$

Also,

${f}^{\prime}(x)=\frac{1}{n}\phantom{\rule{thinmathspace}{0ex}}{x}^{1/n-1}-\frac{1}{x\phantom{\rule{thinmathspace}{0ex}}\mathrm{log}n}.$

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

${x}_{m}=\frac{{n}^{n}}{(\mathrm{log}n{)}^{n}}.$

We have

$f({x}_{m})=\frac{n}{\mathrm{log}n}-\frac{n}{\mathrm{log}n}\phantom{\rule{thinmathspace}{0ex}}(\mathrm{log}n-\mathrm{log}\mathrm{log}n)<0,$

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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