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M26-33

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M26-33  [#permalink]

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New post 16 Sep 2014, 00:26
15
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

46% (01:19) correct 54% (01:49) wrong based on 253 sessions

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Re M26-33  [#permalink]

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New post 16 Sep 2014, 00:26
1
3
Official Solution:


(1) \(\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\).

Simplify to get: \(4^{(x+y)^2-(x-y)^2}=128^{xy}\);

Next, apply \(a^2-b^2=(a-b)(a+b)\) to get: \(4^{4xy}=128^{xy}\);

\(2^{8xy}=2^{7xy}\);

Equate the powers: \(8xy=7xy\);

\(xy=0\), since given that \(x \gt 0\) then \(y=0\), hence \(y^x=0^x=0\). Sufficient.

(2) \(x \ne 1\) and \(x^y=1\). Since \(x \gt 0\) and \(x \ne 1\) then the only case \(x^y=1\) to hold true is when \(y=0\). So, \(y^x=0^x=0\). Sufficient.


Answer: D
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Re: M26-33  [#permalink]

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New post 02 Feb 2016, 08:22
Hi

how is xy=0?; is it because only way 7xy=8xy is when xy=0?
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Re: M26-33  [#permalink]

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New post 02 Feb 2016, 09:11
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M26-33  [#permalink]

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New post 02 Feb 2016, 10:21
I thought one of the two variables must be zero; which will make both sides equal zero and that will prove the equation right
Also, if either x or y is zero that will result zero as the product xy=0 x.0=0

Thanks
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Re: M26-33  [#permalink]

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New post 02 Oct 2016, 09:41
Sorry Bunuel, I'm so confused, how do you apply the difference of squares to get 4xy?
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Re: M26-33  [#permalink]

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New post 02 Oct 2016, 10:30
Wow thank you.
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Re: M26-33  [#permalink]

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New post 31 Aug 2018, 15:18
This is a great question.
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Re: M26-33  [#permalink]

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New post 07 Sep 2018, 02:35
Bunuel wrote:
wmichaelxie wrote:
Sorry Bunuel, I'm so confused, how do you apply the difference of squares to get 4xy?


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


Bunuel I am still confused about this part.

I get (x+y)^2-(x-y)^2=(x+y)(x+y)-(x-y)(x-y)
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Re: M26-33  [#permalink]

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New post 07 Sep 2018, 02:41
fvoci wrote:
Bunuel wrote:
wmichaelxie wrote:
Sorry Bunuel, I'm so confused, how do you apply the difference of squares to get 4xy?


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


Bunuel I am still confused about this part.

I get (x+y)^2-(x-y)^2=(x+y)(x+y)-(x-y)(x-y)


OK.

1. In my reply I applied \(a^2 - b^2 = (a + b)(a - b)\).

2. Else you can just expand (x+y)^2 and (x-y)^2: \((x+y)^2-(x-y)^2=(x^2 + 2xy + y^2)-(x^2-2xy + y^2)=4xy\)
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Re: M26-33  [#permalink]

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New post 07 Sep 2018, 04:09
Bunuel

Perfect! I understand now. Thanks.
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M26-33  [#permalink]

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New post 18 Sep 2018, 07:10
Bunuel wrote:
OK.

1. In my reply I applied \(a^2 - b^2 = (a + b)(a - b)\).

2. Else you can just expand (x+y)^2 and (x-y)^2: \((x+y)^2-(x-y)^2=(x^2 + 2xy + y^2)-(x^2-2xy + y^2)=4xy\)


For everyone else not understanding how to go from \(a^2 - b^2 = (a + b)(a - b)\) and \((x+y)^2-(x-y)^2=(x+y+x-y)(x+y-x+y)=2x*2y=4xy\), it is simply:
\((x+y)^2 - (x-y)^2 =(x+y)+(x-y) * (x+y)-(x-y) = 2x*2y = 4xy\)
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M26-33 &nbs [#permalink] 18 Sep 2018, 07:10
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