What non-zero real value for $x$ satisfies $(7x)^{14}=(14x)^7$?
$\textbf{(A) } \frac17\qquad \textbf{(B) } \frac27\qquad \textbf{(C) } 1\qquad \textbf{(D) } 7\qquad \textbf{(E) } 14$
Taking the seventh root of both sides, we get $(7x)^2=14x$.
Simplifying the LHS gives $49x^2=14x$, which then simplifies to $7x=2$.
Thus, $x=\boxed{\textbf{(B) }\frac{2}{7}}$.