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

Redeem

Exponents

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Senior | Next Rank: 100 Posts
Posts: 31
Joined: Tue Jun 29, 2010 11:02 pm
Location: Honolulu
Thanked: 1 times

Exponents

by bmorgan » Sat Jul 31, 2010 3:56 pm
5^21 x 4^11 = 2x10^n What is the value of n?


I apologize; I forgot to write the answer choices down, but I'm sure that won't stop someone here from getting the right answer. Anyway, I'm more concerned with how to solve the problem than the answer. Thanks, in advance, for your help.
I do not have superior intelligence or faultless looks. I do not captivate a room or run a mile under six minutes. I only succeeded because I was still working long after everyone else went to sleep.

-Greg Evans

Veritas CAT: 680
GMATPrep CAT: 680
Top
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 55
Joined: Tue Jul 27, 2010 8:37 pm
Thanked: 11 times

by KrazyKarl » Sat Jul 31, 2010 4:07 pm
Good question. I learned that, for these, you should break the non-prime bases down to primes to make all of the bases consistent:

5^21 * (2^2)^11 = 2 * (2*5)^n

5^21 * 2^22 = 2 * 2^n * 5^n

Then because the only way to get 5^21 is through that 5 term, n would be 21, and it works because there are 22 n's on the left and n+1 n's on the right, so that would even out too. So I say n = 21. Is that right?
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 31
Joined: Tue Jun 29, 2010 11:02 pm
Location: Honolulu
Thanked: 1 times

by bmorgan » Sat Jul 31, 2010 4:12 pm
Sure is and that makes perfect sense. Thanks, Karl
I do not have superior intelligence or faultless looks. I do not captivate a room or run a mile under six minutes. I only succeeded because I was still working long after everyone else went to sleep.

-Greg Evans

Veritas CAT: 680
GMATPrep CAT: 680
Top
Junior | Next Rank: 30 Posts
Posts: 18
Joined: Thu Aug 05, 2010 3:44 am

by rahul goyal » Tue Aug 10, 2010 1:47 am
KrazyKarl wrote:Good question. I learned that, for these, you should break the non-prime bases down to primes to make all of the bases consistent:

5^21 * (2^2)^11 = 2 * (2*5)^n

5^21 * 2^22 = 2 * 2^n * 5^n

Then because the only way to get 5^21 is through that 5 term, n would be 21, and it works because there are 22 n's on the left and n+1 n's on the right, so that would even out too. So I say n = 21. Is that right?
Thank you karl.it is a good approach.
Top
Senior | Next Rank: 100 Posts
Posts: 34
Joined: Thu Aug 05, 2010 2:39 am

by likithae » Tue Aug 10, 2010 2:26 am
KrazyKarl wrote:Good question. I learned that, for these, you should break the non-prime bases down to primes to make all of the bases consistent:

5^21 * (2^2)^11 = 2 * (2*5)^n

5^21 * 2^22 = 2 * 2^n * 5^n

Then because the only way to get 5^21 is through that 5 term, n would be 21, and it works because there are 22 n's on the left and n+1 n's on the right, so that would even out too. So I say n = 21. Is that right?

thank you........but some what confusing can u explain me clearly............
Top
Post new topic Post Reply
• Page 1 of 1