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PS Please help - exponents

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PS Please help - exponents

by moonlite » Sat Nov 01, 2008 9:00 am
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

OA A
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by cramya » Sat Nov 01, 2008 9:17 am
Moonlinte,

Prime factorize 288 = 2 ^ 5 * 3 ^ 2

x=5 y = 2

Therefore 2 ^ (5-1) * 3 ^ (2-2)
= 2 ^ 4 * 3 ^ 0 (A^0 IS 1- any non zero integer raised to 0 is 1)
= 2 ^4
=16

A)
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by Mani_mba » Sat Nov 01, 2008 9:17 am
(2^x-1) * (3^y-2)

=> (2^x/2) * (3^y/9)

=> (2^x)(3^y)/18

=> 288/18

=> 16

Hence A.
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by moonlite » Sat Nov 01, 2008 9:18 am
Thank you!
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by dmateer25 » Sat Nov 01, 2008 9:21 am
Prime factors of 288 = 2 x 2 x 2 x 2 x2 x 3 x 3 = 2^5 * 3^2

X=5
Y=2

(2^x-1)(3^y-2)
(2^4)(3^0)
16
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