BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

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

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 2898
Joined: Thu Sep 07, 2017 2:49 pm
Thanked: 6 times
Followed by:5 members

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

by Vincen » Mon Oct 19, 2020 9:05 am

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

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

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

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

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

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

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

Answer: B

Source: Magoosh
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8086
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

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


Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Post new topic Post Reply
• Page 1 of 1