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Exponents

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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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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by tlt2372 » Wed Oct 27, 2010 5:38 pm
Thanks!
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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.
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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
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