nx = xⁿ

The solution is 2017^(1/2016) ( or $\sqrt[2016]{2017}$ ). Why? Let's solve this as an equality.

$2017x = x^{2017}$

Divide both sides by $x$ ; left side becomes $2017$ , right side loses one power of $x$ :

$2017 = x^{2016}$

Take the 2016th square root of both sides; left side becomes $\sqrt[2016]{2017}$ , right side becomes $x$ (because $\sqrt[2016]{x^{2016}} = x$ ):

$\sqrt[2016]{2017} = x$

Et voilà, you've found $x$ ; it's $\sqrt[2016]{2017}$ .

Note: $x$ can also be $0$ or $-\sqrt[2016]{2017}$ because $n$ is odd, but the answer was asked to be the absolute value of $x$ ( $|x|$ ) and not being equal to $0$ .

$\begin{aligned} nx&=x^n \\ n &=x^{n-1} \\ \log n &= (n-1) \log x \\ \log x &= \frac {\log n}{n-1} = \log n^{\frac 1{n-1}} \\ \implies x &= n^{\frac 1{n-1}} = \boxed{2017^{\frac 1{2016}}} \end{aligned}$