2006 AMC 10A Problems/Problem 6

Problem

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

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

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 $\mathrm{(B)}$ .

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