Tag: exponential functions

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 $f (x) = x^{\frac{1}{x}}, x > 0 .$ 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 \neq y, x > 0, y > 0 .

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

Solution

Consider the function $f (x) = x^{\frac{1}{x}}$ 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 horizontal line meets the graph twice. Here it gives the approximate pair

(1.471, 7.857) .

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 = t x .

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

\frac{\ln x}{x} = \frac{\ln (t x)}{t x} .

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

(x, y) = (t^{\frac{1}{t - 1}}, t^{\frac{t}{t - 1}}), t > 1 .

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) = ({(\frac{n + 1}{n})}^{n}, {(\frac{n + 1}{n})}^{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} ↑ e and y_{n} ↓ e .

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

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?

September 21, 2026