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if x and y are positive integers, is 4^x(1/3)^y < 1? (1)

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if x and y are positive integers, is 4^x(1/3)^y < 1? (1) [#permalink] New post 09 Sep 2005, 22:54
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if x and y are positive integers, is 4^x(1/3)^y < 1?

(1) y=2x
(2) y=4

Please show work. Thanks a bunch!
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 [#permalink] New post 10 Sep 2005, 03:47
Should be A

y=2x --> 4^x . (1/3)^2x = 4^x. (1/3)^x. (1/3)^x = 4^x. (1/9)^x =
(4/9)^x

4/9 is less than 1 and x a positive integer (bigger than 0) than (4/9)^x is always less than 1. So it is sufficient.

Statement 2 insufficient because x can be much bigger then y and the product may be bigger or smaller than 1.

Answer is A.
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 [#permalink] New post 19 Sep 2005, 14:18
I go with A - 4^x * (1/3)^X

Statement 1 - we know that x and y are integers. Sub. 2x in for y

4^x * (1/3^2x) no matter what number you plug in for x a fraction will result....therefore sufficient.

Statement 2 - No info about X....insufficient
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 [#permalink] New post 19 Sep 2005, 20:30
As both X and Y are positive integers, statement 1 aloneis sufficient

If Y= 2X then we have one variable. Irrespective of the value of X (as its a positive integer), the product will be less than 1.

So the answer is A.
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 [#permalink] New post 20 Sep 2005, 09:16
A as well here. Statement one will give always no.
  [#permalink] 20 Sep 2005, 09:16
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