Bunuel wrote:
What is the least integer p for which \(27^p > 3^{18}\)?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 18
We first re-express 27 as 3^3, and so 27^p = (3^3)^p = 3^3p. Simplifying the inequality, we have:
27^p > 3^18
3^3p > 3^18
3p > 18
p > 6
The least integer greater than 6 is 7.
Answer: B
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