Difference between revisions of "2006 AMC 10A Problems/Problem 6"
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== Problem == | == Problem == | ||
− | What non-zero real value for <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> \mathrm{(A) \ } \frac17\qquad \mathrm{(B) \ } \frac27\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 14 </math> | ||
== Solution == | == Solution == | ||
− | == See | + | Taking the seventh root of both sides, we get <math>(7x)^2=14x</math>. |
− | + | ||
+ | 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>. | ||
+ | |||
+ | == See also == | ||
+ | {{AMC10 box|year=2006|ab=A|num-b=5|num-a=7}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:31, 4 July 2013
Problem
What non-zero real value for satisfies ?
Solution
Taking the seventh root of both sides, we get .
Simplifying the LHS gives , which then simplifies to .
Thus, , and the answer is .
See also
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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