2006 AMC 10A Problems/Problem 6

Revision as of 14:40, 4 August 2006 by Xantos C. Guin (talk | contribs) (added category and link to previous and next problem)


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

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


We first break up 14 into (7x)(2), so that

$(7x)^{14}=(7x\cdot 2)^7\Longrightarrow (7x)^{14}=((7x)^7)(2^7)$

We then divide out $(7x)^7$



We take the 7th root of each side.


$7x=2\Longrightarrow x=\frac{2}{7}, (B)$

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