Exponents

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tlt2372: Senior | Next Rank: 100 Posts; Posts: 48; Joined: Sun Jul 20, 2008 4:45 pm; Location: Tallahassee, FL; Thanked: 2 times

by tlt2372 » Wed Oct 27, 2010 5:01 pm

If (2^x)(3^y) = 288, where x and y are positive integers, then (2^(x-1))(3^(y-2))=?

A)16
B)24
C)48
D)96
E)144

I know you can just pick numbers and get lucky, but is there a shortcut?

Thanks

OA: A

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Source: Beat The GMAT — Problem Solving | Wed Oct 27, 2010 5:01 pm

Taniuca: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Mon Aug 02, 2010 3:25 pm; Thanked: 3 times

by Taniuca » Wed Oct 27, 2010 5:26 pm

PROCESS

2^x. 3^y = 288 find all the factors for 288= 2*2*2*2*2*3*3 that is the same as saying 2^5*3^2

2^x * 3^y = 2^5*3^2
so, x=5 and y=2

Plug the numbers in your second equation

2^ (x-1) * 2 (y-2) =
2^ (5-1) * 2 (2-2) =
2^4 * 2^0=
16 *1= 16

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tlt2372: Senior | Next Rank: 100 Posts; Posts: 48; Joined: Sun Jul 20, 2008 4:45 pm; Location: Tallahassee, FL; Thanked: 2 times

by tlt2372 » Wed Oct 27, 2010 5:38 pm

Thanks!

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limestone: Master | Next Rank: 500 Posts; Posts: 307; Joined: Sun Jul 11, 2010 7:52 pmFollowed by: 1 member; Thanked: 36 times; GMAT Score:640

by limestone » Wed Oct 27, 2010 8:50 pm

Hi,

I have the general rule for this.

If you have :

a^x * b^y = k, and you want to find a^(x-n) * b ^ (y-m), you can divide k to (a^n * b^m)

In this case : 2^x * 3^y = 288, then you wanna find 2^(x-1)* 3^(y-2):

Then 2^(x-1)* 3^(y-2) = 288/ (2^1 * 3^2) = 288/18 = 16

Hope this helps.

"There is nothing either good or bad - but thinking makes it so" - Shakespeare.

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Wed Oct 27, 2010 9:08 pm

tlt2372 wrote:If (2^x)(3^y) = 288, where x and y are positive integers, then (2^(x-1))(3^(y-2))=?

A)16
B)24
C)48
D)96
E)144

I know you can just pick numbers and get lucky, but is there a shortcut?

Thanks

OA: A

Yes, there exists a method to solve this kind of problems.
For variable x and constant n,
(1) a^(x + n) = (a^x)*(a^n)
(2) a^(x - n) = (a^x)/(a^n)

Thus,
[2^(x - 1)][3^(y - 2)] = [(2^x)/(2)][(3^y)/(3^2)] = (2^x)(3^y )/(2*9) = 288/18 =16

The correct answer is A.

Note: Plugging the options work for this particular problem. But if the numbers were large or the question made more complicated, plugging options will become cumbersome and too some extent impossible.

Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

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