Is 8^x 4^y? (1) x y (2) 3x 2y

Question

Is 8^x > 4^y?

(1) x > y
(2) 3x > 2y

BrentGMATPrepNow · Accepted Answer

Target question :    Is 8^x > 4^y?

This is a great candidate for  rephrasing the target question . 
 Aside: We have a free video with tips on rephrasing the target question:  https://www.gmatprepnow.com/module/gmat-data-sufficiency?id=1100

Notice that we can rewrite 8 and 4 with the same BASE to get:  Is (2³)^x > (2²)^y? 
Now apply the power of a power law to get:  Is 2^3x > 2^2y? 
Since 2^2y is always positive, we can safely divide both sides by 2^2y to get:  Is (2^3x)/(2^2y) > 1? 
Simplify to get:  Is 2^(3x - 2y) > 1? 
For 2^(3x -2y) to be greater than 1, the exponent, 3x - 2y, must be greater than 0. 
So, we get:
  REPHRASED target question :    Is 3x - 2y > 0?

At this point, the question can be handled quickly
 
   Statement 1 : x > y   
Can we use this information to answer the  REPHRASED target question ? 
No. 
There are several values of x and y that satisfy statement 1. Here are two: 
Case a: x = 2 and y = 1. In this case 3x - 2y = 3(2) - 2(1) = 4. In other words,  3x - 2y > 0 
Case b: x = -3 and y = -4. In this case 3x - 2y = 3(-3) - 2(-4) = -1. In other words,  3x - 2y < 0 
Since we cannot answer the  REPHRASED target question  with certainty, statement 1 is NOT SUFFICIENT

Statement 2 : 3x > 2y 
Subtract 2y from both sides to get 3x - 2y > 0
PERFECT!! 
This means we can answer the  REPHRASED target question  with certainty. So, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent

MathRevolution · Answer

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is 8^x>4^y?

1) x>y
2) 3x>2y

When you modify the original condition and the question, 3x>2y? is derived from 2^3x>4^2y?, which makes B the answer.

Is 8^x>4^y?

1) x>y
2) 3x>2y

When you modify the original condition and the question, 3x>2y? is derived from 2^3x>4^2y?, which makes B the answer.

mvictor · Answer

the question can be rewritten as:
2^3x > 2^2y?
or the real question:
3x>2y?

tricky one...almost selected D, but remembered that x and y can be negative as well..

1. x>y. if both numbers are positive, then it is sufficient.
if x=-5, y=-6
x>y
but:
3*-5 = -15
2*-6 = -12
3x<2y
since there are different outcomes, 1 alone is insufficient.

2. exactly says what we need - sufficient.

answer is B.

chetan2u · Answer

8^x>4^y.....2^{3x}>2^{2y} ...
so Q basically asks if 3x>2y...

lets see the statements..

1) x > y
x>2y...YES
x=1.1y...NO
Insuff

2) 3x > 2y
sufficient as this is what we are looking for

B

pushpitkc · Answer

8^x = 2^{3x} 
 4^y = 2^{2y} 
In essence, we have been asked to test if  2^{3x} > 2^{2y}

1. x > y 
If x = 2 and y = 1 then  2^6 > 2^2 
But if x = -3 and y = -4, then  2^{-9} < 2^{-8}  (Insufficient)

2. 3x > 2y 
This is enough to determine that  2^{3x} > 2^{2y}  is always true. (Sufficient) (Option B)

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Is 8^x > 4^y? (1) x > y (2) 3x > 2y

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