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M26-33

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Bunuel
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M26-33 [#permalink] New post   16 Sep 2014, 01:26
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46% (02:12) correct 54% (02:31) wrong based on 217 sessions
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If \(x \gt 0\), 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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Bunuel
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Re M26-33 [#permalink] New post   16 Sep 2014, 01:26
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Official Solution:


If \(x \gt 0\), what is the value of \(y^x\)?

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

Simplifying this equation, we get \(4^{(x+y)^2-(x-y)^2}=128^{xy}\).

Expanding and simplifying the exponent of 4 gives \(4^{4xy}=128^{xy}\).

Expressing each side in terms of powers of 2 gives \(2^{8xy}=2^{7xy}\);

Equating the powers, we find \(8xy=7xy\), which implies \(xy=0\).

Since given that \(x > 0\), we can deduce that \(y=0\), and therefore \(y^x=0^x=0\).

Sufficient.

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

Since \(x > 0\) and \(x \ne 1\), the only possibility for \(x^y=1\) to hold true is when \(y=0\). Thus, \(y^x=0^x=0\). Sufficient.


Answer: D
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aqeelnaqvi10
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Re: M26-33 [#permalink] 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] New post   02 Feb 2016, 10:11
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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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Re: M26-33 [#permalink] 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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Bunuel
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Re: M26-33 [#permalink] New post   02 Oct 2016, 10:45
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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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Re: M26-33 [#permalink] New post   02 Oct 2016, 11:30
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Wow thank you.
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Re: M26-33 [#permalink] New post   31 Aug 2018, 16:18
This is a great question.
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Re: M26-33 [#permalink] 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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Bunuel
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Re: M26-33 [#permalink] New post   07 Sep 2018, 03:41
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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] New post   07 Sep 2018, 05:09
Bunuel

Perfect! I understand now. Thanks.
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Re: M26-33 [#permalink] 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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Bunuel
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Re: M26-33 [#permalink] New post   12 Apr 2019, 07:52
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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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Re: M26-33 [#permalink] 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] New post   17 Oct 2019, 03:17
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Hi Bunuel
for example sqrt(2) = ~ 1.414, but if we want to achieve a value exactly = 1 from some(not square root) root of 2 ,
then can't we imagine a situation like (2) ^ 1/(100.xyz....) which can be equal to 1 ?

If that could be the case then its not necessarily that we consider y = 0 in this question.
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Re: M26-33 [#permalink] New post   17 Oct 2019, 03:28
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bhimeshsharma wrote:
Hi Bunuel
for example sqrt(2) = ~ 1.414, but if we want to achieve a value exactly = 1 from some(not square root) root of 2 ,
then can't we imagine a situation like (2) ^ 1/(100.xyz....) which can be equal to 1 ?

If that could be the case then its not necessarily that we consider y = 0 in this question.


No. 2^x = 1 ONLY if x = 0.
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Re: M26-33 [#permalink] New post   21 Apr 2023, 03:01
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M26-33 [#permalink]
21 Apr 2023, 03:01
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