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

(Problem)
(Solution)
 
Line 9: Line 9:
 
Simplifying the LHS gives <math>49x^2=14x</math>, which then simplifies to <math>7x=2</math>.
 
Simplifying the LHS gives <math>49x^2=14x</math>, which then simplifies to <math>7x=2</math>.
  
Thus, <math>x=\frac{2}{7}</math>, and the answer is <math>\mathrm{(B)}</math>.
+
Thus, <math>x=\boxed{\textbf{(B) }\frac{2}{7}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 18:07, 16 December 2021

Problem

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

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

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, $x=\boxed{\textbf{(B) }\frac{2}{7}}$.

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