Solutions to $x^x=1$?

Question

When $x>0$, how many solutions does $x^x=1$ have?

I found that $\lim_{x \to 0} x^x=1$. How does one show that this is the only possible answer?

Answer 1

hint

for $x>0$,

$$x^x=e^{x\ln(x)}=e^0$$

Answer 2

HINT

Note that $x^x$ is not defined for $x=0$ therefore $x=0$ is not a solution.

Of course we have that for $x=1$ we have $1^1=1$.

To show that it is the unique solution let consider

$$f'(x)=x^x(\log x+1)$$

Answer 3

Take logarithms (to any base) to obtain $x\log x=0$ so for $x\gt 0$ (condition given in question - in particular $x\neq 0$) you need $\log x=0$ ie $x=1$.

Answer 4

By the hints above, we have $$ x \ln(x) = 0. $$ As $x >0$ we have $ \ln x =0$ which implies $x=1$ since the logarithmic function is a bijection.

As observed, we cannot have $x=0$ and the case $x<0$ is analogous.

Answer 5

Suppose $x>0$ and $x^x=1$. Raise both sides to the power of $1/x$ to obtain $$(x^x)^{1/x} = 1^{1/x}$$ which implies that $x=1$.

Answer 6

$$y=x^x$$ $$y=e^{x\ln x}$$ we know that $e^0=1$, so: $$x\ln(x)=0$$ $$\ln(1)=0$$ so the solutions are $x=0,1$

although notice that at $x=0$ this is a limit and only exists for $0^+$

Answer 7

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Make the graph of the given function, and see where does the horizontal line $y=1$ cuts it.

Answer 8

Your question is $x^{x}=1$. Now clearly $x=1$ satisfies the above equation because $1^{1}=1$. Again you can solve this above equation by taking $log$ on both sides.

$x^{x}=1$. Therefore $xln(x)=ln(1)$. Now substitute $x=e^{t}$. Therefore $te^{t}=1$. Now here I will introduce the concept of Lambert W Function. That is $W(xe^{x})=x$. Therefore now if I take Lambert W Function on both sides, then we will get $W(te^{t})=W(ln(1))$. Therefore you will get $t=W(ln(1))$. Therefore finally $ln(x)=W(ln(1))$. Therefore, $x=e^{W(ln(1))}$.