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

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

by Max@Math Revolution » Tue Mar 15, 2016 5:48 pm
Is 8^x>4^y?

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


* A solution will be posted in two days.

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by John fran kennedi » Wed Mar 16, 2016 7:41 am
Let me try to solve this
Statement 1:
I first plug 2 for x and 1 for Y
8^2 > 4^1. YES


I plug -2 for x and -3 for y
1/8^-2 > 1/4^-3. NO.
Hence, statement 1 is insufficient

Statement 2
I plug 1 for both x and y
8 > 4. Yes

I plug -1 for x and -2 for y
1/8 > 1/4^2, yes

Statement 2 is sufficient.


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by Brent@GMATPrepNow » Wed Mar 16, 2016 10:01 am
Max@Math Revolution wrote:Is 8^x > 4^y?

1) x > y
2) 3x > 2y
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- ... cy?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
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by Max@Math Revolution » Sat Mar 19, 2016 9:55 pm
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.