Solving equations of type $x^{1/n}=\log_{n} x$ 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 $y = \log_{n} x$ , where $n$ is given and equal for both functions, always have $2$ intersection points. This means, equation  $x^{1/n}= \log_{n} x$ 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 $x^{1/n}= \log_{n} x$ analytically?

 A: They always have two intersection points. Let
$$
f(x)=x^{1/n}-\log_nx.
$$
Then
$$
\lim_{x\to0^+}f(x)=\lim_{x\to+\infty}f(x)=+\infty.
$$
Also,
$$
f'(x)=\frac1n\,x^{1/n-1}-\frac1{x\,\log n}.
$$
It is easy to see that $f'$ has a single zero, which is necessarily a minimum (one can check that $f'$ has the appropriate signs at both sides of this point, which is
$$
x_m=\frac{n^n}{(\log n)^n}.
$$
We have
$$
f(x_m)=\frac{n}{\log n}-\frac{n}{\log n}\,(\log n-\log\log n)<0,
$$
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. 
A: $$\begin{align}\sqrt[n]x&=\log_nx\\\\x&=t^n\end{align}\ \Bigg\}\iff t=\frac n{\ln n}\cdot\ln t\quad;\quad t=e^u\iff e^u=\frac n{\ln n}\cdot u\iff$$
$$(-u)\cdot e^{-u}=-\frac{\ln n}n\iff u=-W\bigg(-\frac{\ln n}n\bigg)\iff x=t^n=(e^u)^n=e^{nu}$$
$$x=\exp\bigg(-n\cdot W\bigg(-\frac{\ln n}n\bigg)\bigg)$$ where W is the Lambert W function.
A: The second solution is explicit in the sense that it can be expressed as
x = [W(z) / z]^n
W noting Lambert function and z = - Ln(n) / n
I use on purpose the notation Ln for natural logarithm.  
What is extremely amazing is that, at least for 4 < n < 13, the value of the second solution varies linearly with n [x = 10.6844 + 1.2821 n ; correlation coefficient greater than 0.99997 !] .  
Sorry for not being able to justify that.  
Thanks for the nice problem and Merry Xmas to everyone.
A: Applying $\log_n$ to both sides yields
$$ \frac1n \log_n x = \log_n \log_n x $$
Noticing that $x$ now appears only as part of $\log_n x$, let's simplify a bit by writing $u=\log_n x$.  Then we have the following equation for $u$:
$$ \frac1n u = \log_n u $$
Using laws of logarithms, we can rewrite the RHS thus:
$$ \frac1n u = \frac{\log u}{\log n} $$
(where the new logs are in any base you like, as long as it's the same in both places, but let's say it's natural logarithms).  Now let's rearrange to make the equation more symmetrical:
$$ \frac{\log u}{u} = \frac{\log n}{n} \tag{1} $$
(The solution $u=n$, corresponding to $x=n^n$, is now obvious, if it wasn't already.)  If you have calculus, then at this point you start analyzing the function $f(t)=\frac{\log t}{t}$; with calculus it's easy to check that it's decreasing for $t\in(1,e]$ and increasing for $t\in[e,\infty)$, which is why you get two solutions: the graph crosses any given horizontal line $y=c$ twice, once on the way down and once on the way up (assuming $c>\frac1e$).  (This analysis is part of exercise 18.34 in Spivak's Calculus.)
If you don't have calculus, you could interpret (1) thus: for any point $(u,\log(u))$ on the graph of the logarithm function, the slope of the line segment from the origin to that point is $\frac{\log u}{u}$.  I expect that one could prove that such a line segment (extended if necessary) meets the graph of log twice (if $u>1$ and $u\ne e$) using the fact that log is a concave function, a fact which has an elementary proof in the form of the AM/GM inequality.  But, alas, I don't see the details right away and I have to run.
A: Let $x^{1/n}=y$, so $\log y=\frac{\log x}{n}$ or $n\log y=\log x$. Moreover
$$
\log_n x = \frac{\log x}{\log n}=\frac{n\log y}{\log n}
$$
Thus your equation becomes
$$
y=\frac{n\log y}{\log n}
$$
(here $\log z$ is always the natural logarithm). So one solution is obvious, $y=n$ (which gives $x=n^n$). We can assume $n>1$, because the equation doesn't make sense for $n=1$.
Now consider the function
$$
f(y)=y\log n-n\log y
$$
defined for $y>0$. It's readily computed that
$$
\lim_{y\to0}f(y)=\infty=\lim_{y\to\infty}f(y).
$$
Moreover
$$
f'(y)=\log n-\frac{n}{y}=\frac{y\log n-n}{y}
$$
which is positive for $y>\frac{n}{\log n}$ and negative for $0<y<\frac{n}{\log n}$. So $y_0=\frac{n}{\log n}$ is the point where $f$ has a minimum and
$$
f(y_0)=\frac{n}{\log n}\log n-n(\log n-\log\log n)=
n(1+\log\log n-\log n)=n\log\frac{e\log n}{n}.
$$
Now $\frac{e\log n}{n}<1$ if and only if $e\log n<n$. consider the function
$$
g(z)=e\log z-z
$$
We have $g'(z)=(e-z)/z$, so $g$ has a maximum at $e$ and $g(e)=e-e=0$. Thus $g(z)<0$ for all $z>0$, $z\ne e$. Therefore $(e\log n)/n<0$ and therefore $f(y_0)<0$.
Note that $n>y_0$, so the second solution for the equation (in the unknown $y$) is between $0$ and $y_0$.
