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

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

Revision as of 01:33, 11 September 2007

Problem

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

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

Solution

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.

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

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

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AMC 10 Problems and Solutions
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