Difference between revisions of "2006 AMC 10A Problems/Problem 6"

Revision as of 01:33, 11 September 2007

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 $(7x)^7$

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

$(7x)^7=2^7$

We take the 7th root of each side.

$\sqrt[7]{(7x)^7}=\sqrt[7]{2^7}$

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

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

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 <math>7x=2\Longrightarrow x=\frac{2}{7}, (B)</math>
-== See Also ==
+== See also ==
+{{AMC10 box|year=2006|ab=A|num-b=5|num-a=7}}
 *[[2006 AMC 10A Problems]]
-*[[2006 AMC 10A Problems/Problem 5|Previous Problem]]
-*[[2006 AMC 10A Problems/Problem 7|Next Problem]]
 [[Category:Introductory Algebra Problems]]