Exponents

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bmorgan: 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 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.

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Source: Beat The GMAT — Problem Solving | Sat Jul 31, 2010 3:56 pm

KrazyKarl: 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?

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bmorgan: 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

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rahul goyal: 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.

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likithae: 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............

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