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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?

Etemon
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  • You can use that $0^0=1$ – Atmos Mar 11 '21 at 13:58
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    @Atmos Isn't $0^0$ undefined? – Etemon Mar 11 '21 at 13:58
  • No $0^0=1$ as $x^x=e^{x \ln \left(x\right)}$ that tends to $e^0=1$ as $x \rightarrow 0$. However $\left(-1\right)^{\left(-1\right)}$ i don't know what it is – Atmos Mar 11 '21 at 13:59
  • Ok. I think $a^{-1}=\frac1a$ so $(-1)^{(-1)}=\frac{1}{-1}=-1$ – Etemon Mar 11 '21 at 14:01
  • @Atmos Did you mean to use a negative rational, but noninteger in your example? – Michael Burr Mar 11 '21 at 14:03
  • @MichaelBurr Yes ! Typing mistake, sorry – Atmos Mar 11 '21 at 14:04
  • @Atmos: Your comment is completely wrong. Limits have nothing to do with the value of $0^0$; read this post. – user21820 Mar 12 '21 at 08:08
  • @user21820 Thanks for the link provided. But considering $x^x$ we can continuously extent $x \mapsto x^x$ by giving the value $1$ when $x=0$ am I right ? – Atmos Mar 12 '21 at 10:52
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    @Atmos: Indeed, you can do so ("continuously exten[d]") for the very specific function $(x↦x^x)$, but it is not relevant here because we are not talking about self exponentiation. $0^0$ should be defined as $1$ for reasons I had given in the linked post, but taking limits in any manner is not a valid reason. – user21820 Mar 12 '21 at 11:25
  • @Atmos $0^0$ is not $1$. $\frac xx$ approches $1$ as $x$ approaches $0$, but that does not mean $\frac00 = 1$ – Some Guy Mar 24 '21 at 04:42
  • @Atmos an equality and an equality of limits are different things – Some Guy Mar 24 '21 at 04:43

1 Answers1

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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 $1$.
Misha Lavrov
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