Find the number of roots of the equation in $\mathbb{R}$

Question

How many roots does the equation $$\\x^{x^x}=(x^x)^x\\$$ have in $\\\mathbb{R}$?

My observations:I observed that $x=-1,1,2$ are its roots.

Are there other roots of this equation?And how we can find them?

Thanks.

Answer 1

You have them all. If $x \lt 0$, $x^x$ is only defined for $x$ integral. If $x$ is integral and less than $-1$, $x^x$ is not integral and $x^{x^x}$ is not defined. Similarly neither side is defined for $x=0$. The only solution less than or equal to zero is $-1$.

For $x \gt 0$, we can write this as $x^{(x^x)}=x^{(x^2)}$ and take natural logs to give $x^x \ln x = x^2 \ln x$. Since $\ln x \neq 0$ unless $x=1$, we can check that case, finding it is a solution, then exclude it and divide by it. This gives $x^x=x^2, x \gt 0 ,x \neq 1$ Taking another log gives $x=2$ as the only remaining solution.

Answer 2

For $ x^{x^x} = x^{x^2} $ the three roots are: 1 ( double root), and 2, found by substitution possible Taylor expansion around root and direct graphing.

Answer 3

$$x^{x^x}=(x^x)^x$$ $$x^{x^x}=x^{x^2}$$

Taking logarithm on both sides,

$${x^x}\,log\, \left\lvert x \right\rvert={x^2}\,log\, \left\lvert x \right\rvert$$

$$({x^x}-{x^2})\,log\,\left\lvert x \right\rvert=0$$

Considering $log\, \left\lvert x \right\rvert=0$, we get $$x=\pm\,1$$

Considering $${x^x}-{x^2}=0$$

$${x^x}={x^2}$$

Taking logarithm on both sides,

$$x\,log\,\left\lvert x \right\rvert=2\,log\,\left\lvert x \right\rvert$$

$$(x-2)\,log\,\left\lvert x \right\rvert=0$$

Therefore, $$x\,=-1\,,\,1,\,2$$