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

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If x and y are positive integers, is (4^x)(1/3)^y < 1? [#permalink] New post 22 Oct 2004, 12:25
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C
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If x and y are positive integers, is (4^x)(1/3)^y < 1?

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

Again I dissagreed with the OA on this one ...
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ans [#permalink] New post 22 Oct 2004, 12:37
answer is C.

both are reqd.

solving y=4, x =2, expression is always < 1

by the way what is OA...............???? 8-)
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 [#permalink] New post 22 Oct 2004, 12:47
OA was C

However I chose A.

If y = 2x, then equation becomes

4^x(1/3)^(2x) or 4^x(1/9)^x or (4/9)^x

To me (4/9)^x seems like it is going to be less than 1 for all positive x? Anyone else care to weight in?
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dude [#permalink] New post 22 Oct 2004, 12:52
what if x is -ve
(4/9)^(-1) > 1
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 [#permalink] New post 22 Oct 2004, 12:53
I think is A


1) sufficient:

y=2x ... (a)

and:

4^x = 2^(2x) ... (b)

replacing (a) and (b) on the question: 2^(2x) * (1/3)^(2x)
= (2*1/3)^(2x)
= (2/3)^(2x)
because x is positive it'll always be <1

2) Insufficient

y =4 and x could be any number, resulting <1 and >1
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 [#permalink] New post 22 Oct 2004, 21:27
I also think it's A...

The question can be written as is (4^x)/(3^y) < 1

From A we get (4^x)/(3^(2x)) Since we know that x is a positive integer, any number we plug in will have the denominator greater than the numerator thus the expression is < 1

From B we only get Y don't know anything about X
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 [#permalink] New post 23 Oct 2004, 07:14
mmm... interesting....

hey tyagel,

Is this question from the OG ?



:?: :? :?
  [#permalink] 23 Oct 2004, 07:14
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