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

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== Problem ==
 
== Problem ==
What non-zero real value for <math>x</math> satisfies <math>(7x)^{14}=(14x)^7</math>?
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What non-zero real value for <math>x</math> satisfies <math>(5x)^{4}=(10x)^3</math>?
  
 
<math> \textbf{(A) } \frac17\qquad \textbf{(B) } \frac27\qquad \textbf{(C) } 1\qquad \textbf{(D) } 7\qquad \textbf{(E) } 14 </math>
 
<math> \textbf{(A) } \frac17\qquad \textbf{(B) } \frac27\qquad \textbf{(C) } 1\qquad \textbf{(D) } 7\qquad \textbf{(E) } 14 </math>

Revision as of 09:56, 2 May 2023

Problem

What non-zero real value for $x$ satisfies $(5x)^{4}=(10x)^3$?

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

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

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