# Difference between revisions of "2006 AMC 10A Problems/Problem 6"

## Problem

What non-zero real value for $x$ satisfies $(7x)^{14}=(14x)^7$?

$\mathrm{(A) \ } \frac17\qquad \mathrm{(B) \ } \frac27\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 14$

## Solution

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, $x=\frac{2}{7}$, and the answer is $\mathrm{(B)}$.