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# M26-33

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:26
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Difficulty:

95% (hard)

Question Stats:

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

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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 $$x^y=1$$

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16 Sep 2014, 01:26
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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.

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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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02 Feb 2016, 10:11
2
aqeelnaqvi10 wrote:
Hi

how is xy=0?; is it because only way 7xy=8xy is when xy=0?

8xy = 7xy;

8xy - 7xy = 7xy - 7xy;

xy = 0.

Hope it's clear.
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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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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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02 Oct 2016, 10:45
1
wmichaelxie wrote:
Sorry Bunuel, I'm so confused, how do you apply the difference of squares to get 4xy?

$$(x+y)^2-(x-y)^2=(x+y+x-y)(x+y-x+y)=2x*2y=4xy$$
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02 Oct 2016, 11:30
Wow thank you.
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31 Aug 2018, 16:18
This is a great question.
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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?

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

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

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

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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07 Sep 2018, 05:09
Bunuel

Perfect! I understand now. Thanks.
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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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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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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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12 Apr 2019, 07:52
1
Mrjack wrote:
this question hurts my brian T_T

how did the original formula get simplified into

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

Operations involving the same bases:
Keep the base, add or subtract the exponent (add for multiplication, subtract for division)
$$a^n*a^m=a^{n+m}$$

$$\frac{a^n}{a^m}=a^{n-m}$$
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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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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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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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# M26-33

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