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\(9^k\cdot 27^{2k}=\)

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by Vincen » Mon Oct 19, 2020 9:05 am
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\(9^k\cdot 27^{2k}=\)

A. \(3^{5+3k}\)

B. \(3^{8k}\)

C. \(3^{11k}\)

D. \(3^{12k}\)

E. \(3^{12k^2}\)

Answer: B

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Re: \(9^k\cdot 27^{2k}=\)

by Scott@TargetTestPrep » Tue Oct 27, 2020 10:09 am
Vincen wrote: ↑
Mon Oct 19, 2020 9:05 am
 \(9^k\cdot 27^{2k}=\)

A. \(3^{5+3k}\)

B. \(3^{8k}\)

C. \(3^{11k}\)

D. \(3^{12k}\)

E. \(3^{12k^2}\)

Answer: B

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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