2006 AMC 10A Problems/Problem 6: Difference between revisions
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<math> \mathrm{(A) \ } \frac17\qquad \mathrm{(B) \ } \frac27\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } 7\qquad \mathrm{(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 == | == Solution == | ||
We first break up 14 into (7x)(2), so that | |||
<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. | |||
<math>\sqrt[7]{(7x)^7}=\sqrt[7]{2^7} | |||
7x=2\Longrightarrow x=\frac{2}{7}, (B)</math> | |||
== See Also == | == See Also == | ||
*[[2006 AMC 10A Problems]] | *[[2006 AMC 10A Problems]] | ||
Revision as of 15:24, 3 July 2006
Problem
What non-zero real value for 
satisfies 
?
Solution
We first break up 14 into (7x)(2), so that
We then divide out 
We take the 7th root of each side.
$\sqrt[7]{(7x)^7}=\sqrt[7]{2^7}
7x=2\Longrightarrow x=\frac{2}{7}, (B)$ (Error compiling LaTeX. Unknown error_msg)
