When Does x^(1/x) Take the Same Value Twice?

We begin with a simple equality:

$$2^{\frac{1}{2}}=4^{\frac{1}{4}}.$$

The problem

Consider the function $f\left(x\right)=x^{\frac{1}{x}},x>0.$ The equality above shows that this function can take the same value at two distinct positive real numbers.

$$x^{\frac{1}{x}}=y^{\frac{1}{y}},x\ne y,x>0,y>0.$$

Prove that infinitely many such pairs exist, and describe all of them.

Solution

Consider the function $f\left(x\right)=x^{\frac{1}{x}}$ on $[1,\infty ).$ It increases until $x=e$ and then decreases toward $1.$ Thus, every horizontal line strictly between $1$ and the maximum intersects the graph twice.

A parametrization of all pairs

We now describe all the solutions. Assume first that $x<y$ and set

$$t=\frac{y}{x}>1.$$

Then

$$y=tx.$$

Taking logarithms of $x^{\frac{1}{x}}=y^{\frac{1}{y}}$ gives

$$\frac{\ln x}{x}=\frac{\ln \left(tx\right)}{tx}.$$

Multiplying by tx, we obtain

$$t\ln x=\ln x+\ln t.$$

Therefore,

$$(t-1)\ln x=\ln t,$$

and hence

$$x=t^{\frac{1}{t-1}}.$$

Since y = tx, it follows that

$$y=t^{\frac{t}{t-1}}.$$

Thus, all solutions with $x<y$ are given by

Conversely, direct substitution shows that every $t>1$ produces a solution. Therefore, this parametrization gives every distinct pair with the smaller number written first. Reversing the two entries gives the solutions with $x>y.$ Since there are infinitely many choices of $t>1,$ there are infinitely many such pairs.

A Few Examples

Choosing $t=\frac{n+1}{n},n=1,2,3,\dots ,$ gives the infinite family of rational solutions

$$(x,y)=({\left(\frac{n+1}{n}\right)}^n,{\left(\frac{n+1}{n}\right)}^{n+1}).$$

Examples from the rational family

n	t	x	y
1	2	2	4
2	3/2	9/4	27/8
3	4/3	64/27	256/81
4	5/4	625/256	3125/1024

Approaching e

These rational solutions approach $e$ from opposite sides. Indeed, if

$$x_n={(1+\frac{1}{n})}^n,y_n={(1+\frac{1}{n})}^{n+1},$$

then

$$x_n\uparrow e\text{and}{y}_n\downarrow e.$$

Thus, the smaller number in each pair approaches $e$ from below, while the larger number approaches $e$ from above.

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Another Same-Value Problem

The function $x^{1/x}$ is not the only familiar-looking function that can take the same value at different inputs. A related question leads to a very different kind of equation.

Continue exploring: When Does x cos(x) Take the Same Value Twice?