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If xy = 1 and x is not equal to y, then what is the value of 7^(1/xy)
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Updated on: 30 Sep 2017, 04:41
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If xy = 1 and x is not equal to y, then what is the value of \((7 ^{(\frac{1}{xy})})^{(\frac{1}{x} \frac{1}{y})}\)? A. \(\frac{1}{49}\) B. \(\frac{1}{7}\) C. 1 D. 7 E. 49
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Originally posted by BloomingLotus on 12 Sep 2017, 09:19.
Last edited by Bunuel on 30 Sep 2017, 04:41, edited 1 time in total.
Edited the question.



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Re: If xy = 1 and x is not equal to y, then what is the value of 7^(1/xy)
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12 Sep 2017, 09:43
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Re: If xy = 1 and x is not equal to y, then what is the value of 7^(1/xy)
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24 Jun 2018, 08:06
BloomingLotus wrote: If xy = 1 and x is not equal to y, then what is the value of \((7 ^{(\frac{1}{xy})})^{(\frac{1}{x} \frac{1}{y})}\)?
A. \(\frac{1}{49}\)
B. \(\frac{1}{7}\)
C. 1
D. 7
E. 49 Will someone post another solution please? I'm having trouble with this one.
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If xy = 1 and x is not equal to y, then what is the value of 7^(1/xy)
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Updated on: 24 Jun 2018, 08:43
Hi, This is my first Solution , so please bear with me. if any doubts, feel free to ask again.
This Question utilizes and plays with exponents way too much and it can get a bit confusing. ALWAYS remember GMAT is all about coming to a conclusion as soon as possible. So with that in mind , let's get started.
X*Y=1 x not equal to y
{7^( 1/ (xy) } ^ [ (1/x)  (1/y) ]
take a minute to observe  separate and merge the possible to a minimized version ..
\((a^b)^c = a^(b*c)\)
Hence, 7^ { [1/ (xy) ] * [ (1/x)  (1/y) ] }
Further take only the parts within the braces and solve it asap ,
={ [1/ (xy) ] * [ (1/x)  (1/y) ] } ={ [1/ (xy) ] * [ (yx)/(xy) ] } A tiny bit of Merging , ={ [ 1*(yx) / (xy) * (xy) ] } ( Still nothing changed from the previous , just rearranged ) ={ [ 1* (1)(x y) / (xy) * (xy) ] } (Now, we can cancel xy in both numerator and denominator, also substitute xy=1 )
= { [ 1* (1) / 1 ] }
Finally we get a simple 1 .
Now back to the question we get 7^ (1) , (i.e) Inverse of 7 = 1/7 .
Hope that helps. Thank you
Originally posted by shyamkarthikk on 24 Jun 2018, 08:41.
Last edited by shyamkarthikk on 24 Jun 2018, 08:43, edited 1 time in total.



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If xy = 1 and x is not equal to y, then what is the value of 7^(1/xy)
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24 Jun 2018, 08:43
msurls wrote: BloomingLotus wrote: If xy = 1 and x is not equal to y, then what is the value of \((7 ^{(\frac{1}{xy})})^{(\frac{1}{x} \frac{1}{y})}\)?
A. \(\frac{1}{49}\)
B. \(\frac{1}{7}\)
C. 1
D. 7
E. 49 Will someone post another solution please? I'm having trouble with this one. Okay lets break down the problem for better understanding you should be aware of the rule \((a^m)^n=a^(m*n)\) , we will be using the same now, compare the question with the standard rule, \((7 ^{(\frac{1}{xy})})^{(\frac{1}{x} \frac{1}{y})}\) we denote a=7, \(m=\frac{1}{xy}\) , \(n=\frac{1}{x} \frac{1}{y}\)as we now from standard that \((a^m)^n=a^(m*n)\), we first try to do m*n\(\frac{1}{xy}*\frac{1}{x} \frac{1}{y}\) => \(\frac{1}{xy}*\frac{yx}{xy}\)=> \(\frac{1}{xy}*\frac{(xy)}{xy}\) As yx=(xy) => cancelling common terms we get \(\frac{1}{xy}*\frac{(xy)}{xy}\) => \(\frac{1}{xy}\)now we have m*n = \(\frac{1}{xy}\) put this in standard form \(7^\frac{1}{xy}\) and since it is given that xy=1 we get \(7^1\), which is = \(\frac{1}{7}\)
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Re: If xy = 1 and x is not equal to y, then what is the value of 7^(1/xy)
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28 Jun 2018, 04:41
doomedcat wrote: msurls wrote: BloomingLotus wrote: If xy = 1 and x is not equal to y, then what is the value of \((7 ^{(\frac{1}{xy})})^{(\frac{1}{x} \frac{1}{y})}\)?
A. \(\frac{1}{49}\)
B. \(\frac{1}{7}\)
C. 1
D. 7
E. 49 Will someone post another solution please? I'm having trouble with this one. Okay lets break down the problem for better understanding you should be aware of the rule \((a^m)^n=a^(m*n)\) , we will be using the same now, compare the question with the standard rule, \((7 ^{(\frac{1}{xy})})^{(\frac{1}{x} \frac{1}{y})}\) we denote a=7, \(m=\frac{1}{xy}\) , \(n=\frac{1}{x} \frac{1}{y}\)as we now from standard that \((a^m)^n=a^(m*n)\), we first try to do m*n\(\frac{1}{xy}*\frac{1}{x} \frac{1}{y}\) => \(\frac{1}{xy}*\frac{yx}{xy}\)=> \(\frac{1}{xy}*\frac{(xy)}{xy}\) As yx=(xy) => cancelling common terms we get \(\frac{1}{xy}*\frac{(xy)}{xy}\) => \(\frac{1}{xy}\)now we have m*n = \(\frac{1}{xy}\) put this in standard form \(7^\frac{1}{xy}\) and since it is given that xy=1 we get \(7^1\), which is = \(\frac{1}{7}\)Thanks man. The only part I don't understand is how \(\frac{1}{x}\)  \(\frac{1}{y}\) = \(\frac{(yx)}{xy}\) This is probably simple math, but I just don't remember it. Any help on this would be appreciated!
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If xy = 1 and x is not equal to y, then what is the value of 7^(1/xy)
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28 Jun 2018, 04:50
Thanks man. The only part I don't understand is how \(\frac{1}{x}\)  \(\frac{1}{y}\) = \(\frac{(yx)}{xy}\) This is probably simple math, but I just don't remember it. Any help on this would be appreciated![/quote] Ok, well that is simple math , but no worries; i did some search and found below link. this will help you understand cross multplication Cheers , no big problem https://www.mathsisfun.com/algebra/crossmultiply.html
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