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

Latest revision as of 19:22, 14 April 2025

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 $49x^2=14x$ . This means that $7x = 2$ , or $x = 0$ .

However, $x$ must be a non-zero value.

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 1: / Line 1: @@
 == 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> \textbf{(A) } \frac17\qquad \textbf{(B) } \frac27\qquad \textbf{(C) } 1\qquad \textbf{(D) } 7\qquad \textbf{(E) } 14 </math>
-== Solution ==
-== See Also ==
+== Solution==
-*[[2006 AMC 10A Problems]]
+Taking the seventh root of both sides, we get <math>(7x)^2=14x</math>.
+Simplifying the LHS gives <math>49x^2=14x</math>. This means that <math>7x = 2</math>, or <math>x = 0</math>.
+However, <math>x</math> must be a non-zero value.
+Thus, <math>x=\boxed{\textbf{(B) }\frac{2}{7}}</math>.
+== See also ==
+{{AMC10 box|year=2006|ab=A|num-b=5|num-a=7}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}