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If x and y are positive integers, is \(4^x – 7^y < 0\)?

OR Is \(4^x-7^y\)?

(1) \(16^{x + 1} < 49^y…………….4^{2(x+1)}<7^{2y}\)
As the numbers are positive integers on both sides, we can take square root of the inequality.
\(4^{x+1}<7^y……….4^x<4^{x+1}<7^y\)
Sufficient

(2) \(x > y\)
If x and y are 2 and 1 or 10 and 1, answer is No.
If x and y are large integers with a difference of just 1, answer is Yes.
\(4^{4}=64……..7^{3}=343\)
Insufficient


A
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If x and y are positive integers, is \(4^x – 7^y < 0\)?

Stat1: \(16^{x + 1} < 49^y\)
or, \(4^{2(x+1)} < 7^{2y}\), as x and y are positive integers, so, 2(x+1) < 2y or, y-x > 1, so the given equation will be true always. Sufficient

Stat2: \(x > y\)
There can be two conditions here, either x = 2 and y =1 then, \(4^x – 7^y > 0\)
and, if x = 200 and y =1 then, \(4^x – 7^y < 0\). Not sufficient

So, I think A. :)
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Bunuel
If x and y are positive integers, is \(4^x – 7^y < 0\)?

(1) \(16^{x + 1} < 49^y\)
since these are positive numbers we can take root therefore
4*4^x+1 <4*x < 7*y
Which is clearly sufficient

(2) \(x > y\)

when x=1 and y=2 eqn is satisfied

however x=100 and y=2 it's not sufficient

Therefore IMO A
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As per statement 1 =

4^x < 7^y/4 so clearly 4^x < 7^y

Statement 1 is sufficient.

Statement 2=
X>Y

x=2, Y=1 Yes
X=4, Y=3 No
Statement 2 is insufficient.

Answer is A.
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