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

Latest revision as of 14:28, 31 May 2023

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

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 == 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> \textbf{(A) } \frac17\qquad \textbf{(B) } \frac27\qquad \textbf{(C) } 1\qquad \textbf{(D) } 7\qquad \textbf{(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 ==
-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>
-We then divide out <math>(7x)^7</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.
+Simplifying the LHS gives <math>49x^2=14x</math>, which then simplifies to <math>7x=2</math>.
-<math>\sqrt[7]{(7x)^7}=\sqrt[7]{2^7}</math>
+Thus, <math>x=\boxed{\textbf{(B) }\frac{2}{7}}</math>.
-<math>7x=2\Longrightarrow x=\frac{2}{7}, (B)</math>
 == See also ==
 {{AMC10 box|year=2006|ab=A|num-b=5|num-a=7}}
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}