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M26-33

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M26-33  [#permalink]

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New post 16 Sep 2014, 01:26
16
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

46% (02:15) correct 54% (02:30) wrong based on 165 sessions

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Re M26-33  [#permalink]

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New post 16 Sep 2014, 01:26
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, 09:22
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, 10:11
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M26-33  [#permalink]

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New post 02 Feb 2016, 11:21
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, 10:41
Sorry Bunuel, I'm so confused, how do you apply the difference of squares to get 4xy?
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New post 02 Oct 2016, 10:45
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Re: M26-33  [#permalink]

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New post 02 Oct 2016, 11:30
Wow thank you.
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Re: M26-33  [#permalink]

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New post 31 Aug 2018, 16:18
This is a great question.
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Re: M26-33  [#permalink]

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New post 07 Sep 2018, 03:35
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, 03:41
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, 05:09
Bunuel

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

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New post 18 Sep 2018, 08:10
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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Re: M26-33  [#permalink]

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New post 15 Dec 2018, 07:13
First, think about statement 2, its quite simple.

Saying x is not equal to 1 and x^y=1, and stem saying x>0
Then just by some intuition, we can deduce that y must 0 and x= positive integer or positive fraction.

Put this values in statement 1. It works.

Both statements are sufficient!
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Re: M26-33  [#permalink]

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New post 12 Apr 2019, 07:18
this question hurts my brian T_T

how did the original formula get simplified into

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

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New post 12 Apr 2019, 07:52
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M26-33  [#permalink]

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New post 12 Apr 2019, 10:08
Bunuel wrote:
If \(x \gt 0\), then what is the value of \(y^x\)?


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

(2) \(x \ne 1\) and \(x^y=1\)



#1
\(\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\)
solve we get

2*(x+y)^2-2*(x-y)^2=7xy

8xy=7xy
we get
xy=0
so it means x or y is 0 ; so value of \(y^x\) will be 0 always; sufficient
#2
\(x \ne 1\) and \(x^y=1\)[/quote]
x is not equal to 1 and x^y =1
so x=0 or say y=0
so value of y^x=0
sufficient
IMO D
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Re M26-33  [#permalink]

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New post 15 Aug 2019, 05:55
I think this is a high-quality question and I agree with explanation. 700L questions seem so easy to solve after viewing the solution. Ha! Anytime, I'm solving a very hard question and it's taking me more than a minute, I just move on. I believe I have missed something and when I check the solution, I'm never surprised to see a 1-5line solution. I hope I improve a whole lot before the exam. Kudos, Bunuel.
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Re: M26-33  [#permalink]

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New post 30 Aug 2019, 06:47
Bunuel wrote:
If \(x \gt 0\), then what is the value of \(y^x\)?


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

(2) \(x \ne 1\) and \(x^y=1\)


Statement (1) can be re-written as

=> 4^(x^2) * 4^(y^2) * 4^(2xy) / 4^(x^2) * 4^(y^2) * 4^(-2xy) = 128^(xy)

=> 4^(2xy)/4^(-2xy) = 128^(xy)

=> 4^(2xy) * 4^(2xy) = 128^(xy)

=> 16^(xy) * 16^(xy) = 128^(xy)

=> 2^(8xy) = 2^(7xy)

=> 2^(xy) = 1

This is true only when xy = 0

Becuase x>0, y = 0. Hence, value of y^x = 0.

Statement 2 is sufficient as well.

Hence, D
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Re: M26-33   [#permalink] 30 Aug 2019, 06:47
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