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

Revision as of 17:07, 16 December 2021

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

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

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

2006 AMC 10A (Problems • Answer Key • Resources)
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

@@ Line 9: / Line 9: @@
 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 ==