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

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== Problem ==
 
== Problem ==
What non-zero real value for <math>\displaystyle x</math> satisfies <math>\displaystyle(7x)^{14}=(14x)^7</math>?
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What non-zero real value for <math>x</math> satisfies <math>(7x)^{14}=(14x)^7</math>?
  
 
<math> \mathrm{(A) \ } \frac17\qquad \mathrm{(B) \ } \frac27\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 14 </math>
 
<math> \mathrm{(A) \ } \frac17\qquad \mathrm{(B) \ } \frac27\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 14 </math>
 
== Solution ==
 
== Solution ==
We first break up 14 into (7x)(2), so that
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Taking the seventh root of both sides, we get <math>(7x)^2=14x</math>.
  
<math>(7x)^{14}=(7x\cdot 2)^7\Longrightarrow (7x)^{14}=((7x)^7)(2^7)</math>
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Simplifying the LHS gives <math>49x^2=14x</math>, which then simplifies to <math>7x=2</math>.
  
We then divide out <math>(7x)^7</math>  
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Thus, <math>x=\frac{2}{7}</math>, and the answer is <math>\mathrm{(B)}</math>.
  
<math>\frac{(7x)^{14}}{(7x)^7}=\frac{((7x)^7)(2^7)}{(7x)^7}</math>
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== See also ==
 
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{{AMC10 box|year=2006|ab=A|num-b=5|num-a=7}}
<math>(7x)^7=2^7</math>
 
 
 
We take the 7th root of each side.
 
 
 
<math>\sqrt[7]{(7x)^7}=\sqrt[7]{2^7}</math>
 
 
 
<math>7x=2\Longrightarrow x=\frac{2}{7}, (B)</math>
 
== See Also ==
 
*[[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]]
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{{MAA Notice}}

Revision as of 11:31, 4 July 2013

Problem

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

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

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

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