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21 Mar 2018, 23:07
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B
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Difficulty:
5%
(low)
Question Stats:
84% (00:27) correct
16%
(00:39)
wrong
based on 152
sessions
History
Date
Time
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Not Attempted Yet
What is the least integer p for which \(27^p > 3^{18}\)?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 18
(A) 6
(B) 7
(C) 8
(D) 9
(E) 18
21 Mar 2018, 23:15
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Bunuel
\(27^p\) can be simplified as \((3^3)^p = 3^{3p}\) because\((a^m)^n = a^{m*n}\)
\(3^{3p} > 3^{18}\) -> \(3p > 18\) -> \(p > 6\)
Therefore, the least value of p for which \(27^p > 3^{18}\) is 7 (Option B)
