This is from the solutions manual to the art of problem solving volume 1 (I have attached the image below). In the equation mentioned, I see that x equals three due to 2^3*3^3 being equal to 6^3;however, I am not seeing why the statement the authors made regarding x having 3 as a factor if it is an integer is true. Can someone please explain how to prove this statement for this equation/the rationale behind it? thanks
On
It is a standard result that if $p$ is prime, then for any positive integers $a$ and $n$, $p | a \iff p | a^n$ (where $b|a$ means $b$ divides $a$). This comes from the fact that if $p|ab$ then either $p|a$ or $p|b$, which you can apply inductively to $p|(a \times a^{n-1})$.
So if $x$ is a positive integer, then $(2x)^x$ is an integer power of $2x$, so then $3|(2x)^x \implies 3|2x$, and since $3 \not | 2$ we must have $3|x$.
If $x$ is an integer, $(2x)^x$ can be prime factored and, by the fundamental theorem of arithmetic, can only have the one factorization $2^33^3$. So, the base number $2x$ must be a multiple of $3$.