If you had an equation where x^x = y where x is a positive number that is not 0 or 1, can there ever be more than 1 value of x?

Question

If $x^x = 1$, both $0$ and $1$ would satisfy $x$. Does this mean $0 = 1$?.

If you had an equation where $x^x = y$ where $x$ is a positive number that is not $0$ or $1$, can there ever be more than $1$ value of $x$?

Answer 1

This is a fun function to play with. I had a minor obsession with it before i took calculus. Yes there are y-values with more than one x-value located on $(0,1)$ Note the global min of this as a real function is when $x=e^{-1}$

Answer 2

For any point $x_1$ in the range $[0,1]$, there is another point $x_2$ in the same range such that: $$x_1^{x_1}= x_2^{x_2}$$

Answer 3

To answer your first question, $0^0$ is NOT equal to $1$; instead, it is undefined. Even if $0$ did satisfy the equation, it would not imply that $1=0$. $-1^2=1$ and $1^1=1$, but this does not imply that $-1=1$. They are merely two distinct solutions to the equation $x^2=1$

To answer your second question, it is possible for, for example, $y=0.75$