Number of solutions of $x(x^x-x)=0$

algebra-precalculuscontest-math

How many solutions the equation $x(x^x-x)=0$ has in real numbers set?
$$1)\text{zero}\quad\quad\quad\quad\quad2)\text{one}\quad\quad\quad\quad\quad3)\text{two}\quad\quad\quad\quad\quad4)\text{three}\quad\quad\quad\quad\quad5)\text{infinity}$$

Here is my work:

To solve this problem I considered two cases: $x=0$ or $x^x-x=0$. for $x=0$ I think $0(0^0-0)$ is undefined in real numbers( because we have $0^0$). so we should solve $x^x=x$ however I'm not sure how we should solve this but it is easy to see $x=\pm1$ are the roots. so I think the equation has two roots in real numbers.

Is my answer right?

Best Answer

The answer is either "two roots" or "three roots" depending on if you consider $0^0$ to be $1$ or undefined, and both are common, so it's a poorly written question.

Ignoring the $x=0$ case, you are right that $x^x-x=0$ is only possible when $x=\pm1$. We can divide through by $x$ to get $x^{x-1} = 1$. Some casework, now:

  • One way to have $x^y = 1$ is if $y=0$; here, that means $x-1=0$, so $x=1$.
  • If the power is nonzero, then only powers of $\pm 1$ can be $1$.
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