\(9^k\cdot 27^{2k}=\)
A. \(3^{5+3k}\)
B. \(3^{8k}\)
C. \(3^{11k}\)
D. \(3^{12k}\)
E. \(3^{12k^2}\)
Answer: B
Source: Magoosh
\(9^k\cdot 27^{2k}=\)
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Solution:
Rewriting both 9 and 27 as powers of 3, we have:
(3^2)^k * (3^3)^(2k) = 3^(2k) * 3^(6k) = 3^(8k)
Answer: B
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