2006 AMC 10A Problem 6

Below is the professionally curated solution for Problem 6 of the 2006 AMC 10A, 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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Concepts:exponent algebraic manipulation

Difficulty rating: 1190

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

$\dfrac{1}{7}$

$\dfrac{2}{7}$

$1$

$7$

$14$

Solution:

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

Thus, the correct answer is B.

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