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Niece teaser on Exponents

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DestinyChild
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Niece teaser on Exponents [#permalink] New post   10 Jan 2010, 18:58
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (02:13) correct 0% (00:00) wrong based on 2 sessions
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Got me stumped for no good reason.



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Re: Niece teaser on Exponents [#permalink] New post   10 Jan 2010, 19:27
2^4 = 16 is the answer.

Basically (2^x)/(2^y) = 2^(x-y) and then, there's some algebra which I am sure, you can figure it out yourselves.
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Re: Niece teaser on Exponents [#permalink] New post   10 Jan 2010, 19:58
x and y both equal to 1 or -1.

Plug the numbers in and you end up with 2^4=16 in both cases.
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Re: Niece teaser on Exponents [#permalink] New post   12 Jan 2010, 15:50
yup! I had... just put it out there...!
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dmetla
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Re: Niece teaser on Exponents [#permalink] New post   12 Jan 2010, 17:04
if u solve it its 2^4=16
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Re: Niece teaser on Exponents [#permalink] New post   13 Jan 2010, 11:35
I still dont get. Can you please fully explain...thanks.
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Re: Niece teaser on Exponents [#permalink] New post   14 Jan 2010, 02:31
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cameraknox wrote:
I still dont get. Can you please fully explain...thanks.

Sure.

Here is what you need to know:
A. \((x+y)^2 = x^2 +2xy +y^2\)
B. \((x-y)^2 = x^2 -2xy +y^2\)
C. As BarneyStinson has mentioned you need to know that \((2^x)/(2^y) = 2^{x-y}\)

Now solve the problem using the formulas provided above.

\(2^{x^2 +2xy +y^2}/2^{x^2 -2xy +y^2}\);
\(2^{x^2 +2*1 +y^2}/2^{x^2 -2*1 +y^2}\); (Remember xy = 1)

By using formula C you get \(2^{x^2 +2 +y^2 -x^2 -(-2) -y^2}\)
And finally after canceling \(x^2\) and \(y^2\) out, you get \(2^{2-(-2)}\)
What in turn is equal to \(2^4\)or just 16

I hope this explanation helps you.
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Re: Niece teaser on Exponents [#permalink] New post   28 Jan 2010, 11:52
Thanks DestinyChild. I enjoyed a lot this problem.

Since \(\frac{2^{(x+y)^2}} {2^{(x-y)^2}}\) then, you can get two quadratic equations from the exponents: \({(x+y)^2 = x^2 + 2xy + y^2}\) and \({(x-y)^2 = x^2 - 2xy + y^2}\)

Therefore, \(X^2\) and \(y^2\) both in the numerator and denominator should be canceled, which is equivalent to \(\frac{2^{2xy}} {2^{-2xy}}\). This implies that \(\frac{2^{2xy}} {\frac{1} {2^{2xy}}}\).

Because \(xy = 1\) you can calculate the exponents values to get \(\frac{2^2} {\frac{1} {2^2}}\) or \(\frac{4} {\frac{1} {4}}\). Finally, applying the double C you'll get 16.

Hope this helps :P
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Re: Niece teaser on Exponents [#permalink] New post   29 Jan 2010, 10:24
glad you liked it...

i bet you can still get this wrong on any day... have an argument with your gf\wife and try it... :)
or try it after shoveling 28" of snow... :) :)
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johnhillescobar
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Re: Niece teaser on Exponents [#permalink] New post   29 Jan 2010, 12:17
Of course DestinyChild...it's possible!!!! :-D
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shrivastavarohit
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Re: Niece teaser on Exponents [#permalink] New post   14 Apr 2010, 02:27
Pretty late however this is simpler, please make a lots of noise in case approach is not correct.

2^(x+y)^2/2^(x-y)^2

Represent this eq. in non fraction format since base is 2 (same) the eq. should look as below.


xy = 1;

=2^(x^2 + y^2 + 2xy) - (x^2+y^2-2xy)
=2^4xy

Since xy=1

=2^4 = 16
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Re: Niece teaser on Exponents [#permalink] New post   14 Apr 2010, 02:58
mrblack wrote:
x and y both equal to 1 or -1.

Plug the numbers in and you end up with 2^4=16 in both cases.


Not necessarily, the question doesn't state that they are both integers.
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Re: Niece teaser on Exponents [#permalink] New post   19 May 2010, 20:11
nickk wrote:
mrblack wrote:
x and y both equal to 1 or -1.

Plug the numbers in and you end up with 2^4=16 in both cases.


Not necessarily, the question doesn't state that they are both integers.



it doesn't matter if they are integers or not ....the final answer doesn't change



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Re: Niece teaser on Exponents [#permalink]
19 May 2010, 20:11
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