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

m (See Also)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
What non-zero real value for <math>\displaystyle x</math> satisfies <math>\displaystyle(7x)^{14}=(14x)^7</math>?
+
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
+
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>
+
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>  
+
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>
 
 
<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 ==
 
== See also ==
 
{{AMC10 box|year=2006|ab=A|num-b=5|num-a=7}}
 
{{AMC10 box|year=2006|ab=A|num-b=5|num-a=7}}
 
*[[2006 AMC 10A Problems]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 10:31, 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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS