2006 AMC 10A Problems/Problem 6

Problem

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 $49x^2=14x$ . This means that $7x = 2$ , or $x = 0$ .

However, $x$ must be a non-zero value.

Thus, $x=\boxed{\textbf{(B) }\frac{2}{7}}$ .

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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