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If xy = 1 and x is not equal to y, then what is the value of 7^(1/x-y)

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If xy = 1 and x is not equal to y, then what is the value of 7^(1/x-y)  [#permalink]

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If xy = 1 and x is not equal to y, then what is the value of \((7 ^{(\frac{1}{x-y})})^{(\frac{1}{x}- \frac{1}{y})}\)?


A. \(\frac{1}{49}\)

B. \(\frac{1}{7}\)

C. 1

D. 7

E. 49

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/x-y)  [#permalink]

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New post 12 Sep 2017, 09:43
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Kudos if it helps. :-)
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Re: If xy = 1 and x is not equal to y, then what is the value of 7^(1/x-y)  [#permalink]

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New post 24 Jun 2018, 08:06
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BloomingLotus wrote:
If xy = 1 and x is not equal to y, then what is the value of \((7 ^{(\frac{1}{x-y})})^{(\frac{1}{x}- \frac{1}{y})}\)?


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/x-y)  [#permalink]

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New post 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/ (x-y) } ^ [ (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/ (x-y) ] * [ (1/x) - (1/y) ] }

Further take only the parts within the braces and solve it asap ,

={ [1/ (x-y) ] * [ (1/x) - (1/y) ] }
={ [1/ (x-y) ] * [ (y-x)/(xy) ] }
A tiny bit of Merging ,
={ [ 1*(y-x) / (x-y) * (xy) ] } ( Still nothing changed from the previous , just re-arranged )
={ [ 1* (-1)(x -y) / (x-y) * (xy) ] } (Now, we can cancel x-y 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/x-y)  [#permalink]

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New post 24 Jun 2018, 08:43
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1
msurls wrote:
BloomingLotus wrote:
If xy = 1 and x is not equal to y, then what is the value of \((7 ^{(\frac{1}{x-y})})^{(\frac{1}{x}- \frac{1}{y})}\)?


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, \((7 ^{(\frac{1}{x-y})})^{(\frac{1}{x}- \frac{1}{y})}\)

we denote a=7, \(m=\frac{1}{x-y}\) , \(n=\frac{1}{x}- \frac{1}{y}\)

as we now from standard that \((a^m)^n=a^(m*n)\), we first try to do m*n

\(\frac{1}{x-y}*\frac{1}{x}- \frac{1}{y}\)

=> \(\frac{1}{x-y}*\frac{y-x}{xy}\)

=> \(\frac{1}{x-y}*\frac{-(x-y)}{xy}\) As y-x=-(x-y)

=> cancelling common terms we get \(\frac{1}{x-y}*\frac{-(x-y)}{x-y}\) => \(\frac{-1}{xy}\)

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/x-y)  [#permalink]

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New post 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 \((7 ^{(\frac{1}{x-y})})^{(\frac{1}{x}- \frac{1}{y})}\)?


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, \((7 ^{(\frac{1}{x-y})})^{(\frac{1}{x}- \frac{1}{y})}\)

we denote a=7, \(m=\frac{1}{x-y}\) , \(n=\frac{1}{x}- \frac{1}{y}\)

as we now from standard that \((a^m)^n=a^(m*n)\), we first try to do m*n

\(\frac{1}{x-y}*\frac{1}{x}- \frac{1}{y}\)

=> \(\frac{1}{x-y}*\frac{y-x}{xy}\)

=> \(\frac{1}{x-y}*\frac{-(x-y)}{xy}\) As y-x=-(x-y)

=> cancelling common terms we get \(\frac{1}{x-y}*\frac{-(x-y)}{x-y}\) => \(\frac{-1}{xy}\)

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 \(\frac{1}{x}\) - \(\frac{1}{y}\) = \(\frac{(y-x)}{xy}\)

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/x-y)  [#permalink]

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New post 28 Jun 2018, 04:50
Thanks man. The only part I don't understand is how \(\frac{1}{x}\) - \(\frac{1}{y}\) = \(\frac{(y-x)}{xy}\)

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/cross-multiply.html
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If xy = 1 and x is not equal to y, then what is the value of 7^(1/x-y) &nbs [#permalink] 28 Jun 2018, 04:50
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