The solution gets pretty messy when using a basic text editor. So, here's a graphic insteadRudy414 wrote:If xy=1, what is the value of (2^(x+y)^2)/(2^(x-y)^2)?
2
4
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16
32
Answer = D
Cheers,
Brent
Find the Value
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Brent@GMATPrepNow
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Answer = D
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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Atekihcan
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Easiest approach to solve this problem is to note that the answer choices are independent of x and y. So, we can pick any value for x and y such that xy = 1.
Obvious numbers to pick are x = 1 and y = 1
So, (2^(x + y)^2)/(2^(x - y)^2) = (2^(1 + 1)^2)/(2^(1 - 1)^2) = (2^(2^2))/(2^(0^2)) = (2^4)/(2^0) = 16/1 = 16
Answer : D
Obvious numbers to pick are x = 1 and y = 1
So, (2^(x + y)^2)/(2^(x - y)^2) = (2^(1 + 1)^2)/(2^(1 - 1)^2) = (2^(2^2))/(2^(0^2)) = (2^4)/(2^0) = 16/1 = 16
Answer : D
Last edited by Atekihcan on Wed May 15, 2013 4:33 am, edited 1 time in total.
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Brent@GMATPrepNow
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Excellent approach, Atekihcan!
Mine was overly algebraic.
Cheers,
Brent
Mine was overly algebraic.
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Thank you. I like both approaches. I tend to choose the algebriac way over the pick numbers way just because by doing that I get better understanding of the concept. But the pick numbers makes a lot of sense too.
One question though.
(2^(x + y)^2)/(2^(x - y)^2) = (2^(1 + 1)^2)/(2^(1 - 1)^2) = ((2^2)^2)/((2^0)^2) = (2^4)/(1^2) = 2^4 = 16
As far as order of operations goes, wouldn't this be (2^(2^2))/(2^(0^2))? For this example,I guess it doesn't matter; either way you get 16. But wouldn't you square the exponent before you square the base?
One question though.
(2^(x + y)^2)/(2^(x - y)^2) = (2^(1 + 1)^2)/(2^(1 - 1)^2) = ((2^2)^2)/((2^0)^2) = (2^4)/(1^2) = 2^4 = 16
As far as order of operations goes, wouldn't this be (2^(2^2))/(2^(0^2))? For this example,I guess it doesn't matter; either way you get 16. But wouldn't you square the exponent before you square the base?
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Atekihcan
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Yes, I also prefer algebraic approach to always sticking with plugging numbers. But that doesn't mean I always stick with Algebra. Plugging numbers and other non-conventional methods has their own place. For example, this one.Rudy414 wrote:Thank you. I like both approaches. I tend to choose the algebriac way over the pick numbers way just because by doing that I get better understanding of the concept. But the pick numbers makes a lot of sense too.
Yes. My mistake.Rudy414 wrote:As far as order of operations goes, wouldn't this be (2^(2^2))/(2^(0^2))? For this example,I guess it doesn't matter; either way you get 16. But wouldn't you square the exponent before you square the base?
Editing it right away.
PS : Although the way you have posted the problem "(2^(x+y)^2)" can be interpreted as either [(2^(x+y))^2] or [2^((x+y)^2)]
