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2006 AMC 10A Problems/Problem 6

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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) \ } 1...

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)}.

See also

2006 AMC 10A (ProblemsResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
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