Difference between revisions of "2006 AMC 10A Problems/Problem 6"
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== Problem == | == Problem == | ||
| − | What non-zero real value for <math>x</math> satisfies <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> \textbf{(A) } \frac17\qquad \textbf{(B) } \frac27\qquad \textbf{(C) } 1\qquad \textbf{(D) } 7\qquad \textbf{(E) } 14 </math> | ||
| − | == Solution == | + | == Solution== |
Taking the seventh root of both sides, we get <math>(7x)^2=14x</math>. | Taking the seventh root of both sides, we get <math>(7x)^2=14x</math>. | ||
| − | Simplifying the LHS gives <math>49x^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>. | Thus, <math>x=\boxed{\textbf{(B) }\frac{2}{7}}</math>. | ||
| − | |||
== See also == | == See also == | ||
{{AMC10 box|year=2006|ab=A|num-b=5|num-a=7}} | {{AMC10 box|year=2006|ab=A|num-b=5|num-a=7}} | ||
Latest revision as of 20:22, 14 April 2025
Problem
What non-zero real value for 
satisfies 
?
Solution
Taking the seventh root of both sides, we get 
.
Simplifying the LHS gives 
. This means that 
, or 
.
However, must be a non-zero value.
Thus, 
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
