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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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Intern
Joined: 07 Dec 2016
Posts: 40
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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Updated on: 30 Sep 2017, 03:41
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35% (medium)

Question Stats:

70% (01:40) correct 30% (01:56) wrong based on 76 sessions

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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, 08:19.
Last edited by Bunuel on 30 Sep 2017, 03:41, edited 1 time in total.
Edited the question.
Manager
Joined: 12 Feb 2017
Posts: 70
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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12 Sep 2017, 08:43
1
Kudos if it helps.
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Manager
Status: Studying SC
Joined: 04 Sep 2017
Posts: 120
GPA: 3.6
WE: Sales (Computer Software)
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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24 Jun 2018, 07:06
1
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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Intern
Joined: 19 May 2018
Posts: 1
Location: India
Concentration: Marketing, General Management
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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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Updated on: 24 Jun 2018, 07:43
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, 07:41.
Last edited by shyamkarthikk on 24 Jun 2018, 07:43, edited 1 time in total.
Manager
Joined: 24 Jun 2013
Posts: 132
Location: India
Schools: ISB '20
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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24 Jun 2018, 07:43
3
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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28 Jun 2018, 03:41
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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Manager
Joined: 24 Jun 2013
Posts: 132
Location: India
Schools: ISB '20
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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28 Jun 2018, 03:50
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, 03:50
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