Simplifying exponents

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G- Unit30: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Sun Mar 03, 2013 7:10 pm

by G- Unit30 » Tue Mar 05, 2013 4:17 pm

Hi, I had trouble with this Question from a GMATprep practice exam:

2^5 + 2^5 + 3^5 +3^5 + 3^5 =?

Im not sure what the correct answer was as I can't review gmatprep solutions.

Thanks!

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Source: Beat The GMAT — Problem Solving | Tue Mar 05, 2013 4:17 pm

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue Mar 05, 2013 4:32 pm

G- Unit30 wrote:Hi, I had trouble with this Question from a GMATprep practice exam:

2^5 + 2^5 + 3^5 +3^5 + 3^5 =?

First, notice that K + K = 2K
Using the same logic, 2^5 + 2^5 = 2(2^5) = (2^1)(2^5) = 2^6

Similarly, notice that M + M + M = 3M
Using the same logic, 3^5 + 3^5 + 3^5 = 3(3^5) = (3^1)(3^5) = 3^6

So, 2^5 + 2^5 + 3^5 + 3^5 + 3^5 = 2^6 + 3^6

Aside: For some reason, I remember the question differently.
I think it's (2^5 + 2^5)(3^5 + 3^5 + 3^5) = ?
If that's the correct question, then we can evaluate it as follows:
2^5 + 2^5)(3^5 +3^5 + 3^5) = (2^6)(3^6)
= 6^6 [we're using the rule that says (x^n)(y^n) = (xy)^n]

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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DCS80: Senior | Next Rank: 100 Posts; Posts: 33; Joined: Mon Dec 03, 2012 6:08 pmFollowed by: 2 members; Thanked: 2 times

by DCS80 » Wed Mar 06, 2013 8:20 am

First, notice that K + K = 2K
Using the same logic, 2^5 + 2^5 = 2(2^5) = (2^1)(2^5) = 2^6

Similarly, notice that M + M + M = 3M
Using the same logic, 3^5 + 3^5 + 3^5 = 3(3^5) = (3^1)(3^5) = 3^6

can you explain how the first and second both equate to ^6? (given two factors versus three?)

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed Mar 06, 2013 8:30 am

DCS80 wrote:First, notice that K + K = 2K
Using the same logic, 2^5 + 2^5 = 2(2^5) = (2^1)(2^5) = 2^6

Similarly, notice that M + M + M = 3M
Using the same logic, 3^5 + 3^5 + 3^5 = 3(3^5) = (3^1)(3^5) = 3^6

can you explain how the first and second both equate to ^6? (given two factors versus three?)

You bet.

1 thing + 1 thing = 2things
2^5 + 2^5 = 2(2^5)
At this point, we can replace 2 with 2^1 to get: (2^1)(2^5)
Apply the product law to get: 2^6

Aside: Product Law: (x^a)(x^b) = x^(a+b)

Similarly, 1 thing + 1 thing + 1 thing = 3things
So, 3^5 + 3^5 + 3^5 = 3(3^5)
Replace 3 with 3^1 to get: (3^1)(3^5)
Apply the product law to get: 3^6

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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DCS80: Senior | Next Rank: 100 Posts; Posts: 33; Joined: Mon Dec 03, 2012 6:08 pmFollowed by: 2 members; Thanked: 2 times

by DCS80 » Wed Mar 06, 2013 9:38 am

thx...sorry, was not computing earlier!

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