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If 2^(x + y) = 4^8, then

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jjjinapinch Senior | Next Rank: 100 Posts Default Avatar
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Posted:
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If 2^(x + y) = 4^8, then

Post Tue Aug 08, 2017 7:52 am
If 2^(x + y) = 4^8, then what is the value of y?
(1) x^2 = 81
(2) x - y = 2

Official Guide question
Answer: B

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Top Reply
Post Tue Aug 08, 2017 8:07 am
jjjinapinch wrote:
If 2^(x + y) = 4^8, then what is the value of y?
(1) x² = 81
(2) x - y = 2

Official Guide question
Answer: B
Target question: What is the value of y?

Given: 2^(x + y) = 4^8
Rewrite right side with base of 2.
So, first replace 4 with 2² to get: 2^(x + y) = (2²)^8
Simplify right side: 2^(x + y) = 2^16
We can conclude that x + y = 16

Statement 1: x² = 81
This tells us that EITHER x = 9 OR x = -9
Let's test each case.
Case a: If x = 9, then we can take x + y = 16 and replace x with 9 to get: 9 + y = 16. When we solve this equation for y, we get y = 7
Case b: If x = -9, then we can take x + y = 16 and replace x with -9 to get: -9 + y = 16. When we solve this equation for y, we get y = 25
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x - y = 2
Since we also know that x + y = 16 , we now have a system of 2 variables with equations, which we COULD solve for y (but we won't do because we don't want to waste valuable time).
Since we COULD answer the target question with certainty, statement 2 is SUFFICIENT

Aside: For "fun" let's solve the following system:
x + y = 16
x - y = 2

If we SUBTRACT the bottom equation from the top equation, we get: 2y = 14, which means y = 7
Voila!

Answer: B

Cheers,
Brent

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