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

Question

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 ...

bigpapi · Answer

answer is C. both are reqd. solving y=4, x =2, expression is always < 1 by the way what is OA...............???? 8-)

tyagel · Answer

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?

bigpapi · Answer

what if x is -ve
 (4/9)^(-1) > 1

GMATPIPO · Answer

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

nzgmat · Answer

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

GMATPIPO · Answer

mmm... interesting.... hey tyagel, Is this question from the OG ? :?:

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

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