For which $n$, will $n^x$ and $x$ be equal for just $1$ value of $x$

Question

So, I was playing around in desmos graphing calculator with functions of type $n^x$. I began by plotting $y=2^x$ and $y=x$. I saw that $2^x$ never equals x(kinda obvious but I am still a high schooler).

I then changed $2$ to a variable $n$, changing its value I found that almost all values of $n$ for $n<1.5$ gave 2 solutions to $n^x=x$.

I started narrowing the values of $n$ such that $n^x=x$ gives only 1 solution, or 2 solutions both equal(kinda the same though). Desmos just gave up at $n=1.444667861$.

Of course, it doesn't make sense for such a value to be rational. Hence, I was looking for a method to find the value of this constant, or atleast a method to approximate it using mathematics. But as you know I am just a high schooler.

[P.S.: This question was born out of curiousity, with no relation with my school]

Answer 1

You are trying to solve $x=a^x$ or, using logarithms,

$$\log a=\frac{\log x}x.$$

Looking at the RHS, you see that this function has a single maximum, which occurs where

$$\frac xx-\log x=0.$$

From this you draw the value of $a$ leading to a double root,

$$a=e^x=e^{1/e}.$$

Answer 2

Two graphs can only be tangent (= "just barely touching") if their derivatives are equal at a point where they intersect. The derivative of $n^x$ is $\log n \cdot n^x$, and the derivative of $x$ is $1$; these are only equal when $n^x=1/\log{n}$, or $x=-\log\log n / \log n$. That $x$-value only represents an intersection point if $$n^x=x \implies -\frac{\log\log n}{\log n} = \frac{1}{\log n} \implies \log\log n = -1$$... that is, when $n=e^{e^{-1}}=e^{1/e}.$

In answer to your original question, though, you also have the two graphs intersecting at exactly one point (not tangent) for all $0 \le n \le 1$.