2006 AMC 10A Problem 6

Attempt Problem 6 of the 2006 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10A solutions, or check the answer key.

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6.

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

17\dfrac{1}{7}

27\dfrac{2}{7}

11

77

1414

Answer: B
Concepts:exponentalgebraic manipulation
Difficulty rating: 1190
Solution:

Taking the seventh root of both sides gives (7x)2=14x,(7x)^2 = 14x, so 49x2=14x.49x^2 = 14x. Since x0,x \neq 0, divide by xx to get 49x=14,49x = 14, hence x=27.x = \dfrac{2}{7}.

Thus, the correct answer is B.

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