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# M07-16

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Math Expert
Joined: 02 Sep 2009
Posts: 49993

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16 Sep 2014, 00:34
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Difficulty:

65% (hard)

Question Stats:

61% (01:35) correct 39% (02:14) wrong based on 120 sessions

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If $$x$$ and $$y$$ are positive integers and $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$, then what is the value of $$y-x$$?

A. -18
B. -17
C. 1
D. 17
E. 18

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Math Expert
Joined: 02 Sep 2009
Posts: 49993

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16 Sep 2014, 00:34
Official Solution:

If $$x$$ and $$y$$ are positive integers and $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$, then what is the value of $$y-x$$?

A. -18
B. -17
C. 1
D. 17
E. 18

$$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$;

Cross-multiply: $$28^{18y}=7^x*8^{12}$$;

Factorize: $$2^{36y}*7^{18y}=7^x*2^{36}$$;

Equate the powers of 2 on both sides: $$36y=36$$ --> $$y=1$$;

Equate the powers of 7 on both sides: $$18*1=x$$ --> $$x=18$$;

$$y-x=1-18=-17$$.

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Joined: 19 Aug 2014
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19 Jan 2015, 04:13
Had problems with the website (became irresponsive) during this question causing me to spend almost 12 mins in it. Couldn't finish the test after.
Math Expert
Joined: 02 Sep 2009
Posts: 49993

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19 Jan 2015, 04:14
Ivan90 wrote:
Had problems with the website (became irresponsive) during this question causing me to spend almost 12 mins in it. Couldn't finish the test after.

Thank you for reporting this. We are looking into it.
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Intern
Joined: 09 May 2015
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26 Jun 2015, 15:20
I'm a little rusty, how do you cross multiply to get everything to be a divisor of 1 to being the number alone to the exponent. That is where I struggled with this problem. Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 49993

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27 Jun 2015, 03:43
ConorMcGrath1 wrote:
I'm a little rusty, how do you cross multiply to get everything to be a divisor of 1 to being the number alone to the exponent. That is where I struggled with this problem. Thanks.

$$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$;

$$\frac{1}{7^x}*\frac{1}{8^{12}}=\frac{1}{28^{18y}}$$;

Cross-multiply: $$28^{18y}=7^x*8^{12}$$.
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Joined: 08 Oct 2014
Posts: 5
Location: United States (TX)
Concentration: Entrepreneurship, Strategy
GMAT 1: 680 Q47 V35
GPA: 3.25
WE: Account Management (Mutual Funds and Brokerage)

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22 Dec 2015, 07:16
Not sure I am following when you equate the exponents of 7. Would that not be 18y=x?

Thanks for the help.
Math Expert
Joined: 02 Sep 2009
Posts: 49993

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22 Dec 2015, 07:22
mccalleg wrote:
Not sure I am following when you equate the exponents of 7. Would that not be 18y=x?

Thanks for the help.

Yes, but from the line just above it we established that y = 1, thus 18*1 = x.
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Joined: 05 Jul 2016
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17 Jul 2016, 10:43
Hello,

How does the left hand side of this equation factor? ie the 28^18y

Sorry, a bit rusty on the the exponent rules. Thanks in advance.
Math Expert
Joined: 02 Sep 2009
Posts: 49993

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17 Jul 2016, 10:49
abhilash53 wrote:
Hello,

How does the left hand side of this equation factor? ie the 28^18y

Sorry, a bit rusty on the the exponent rules. Thanks in advance.

$$28^{18y}$$;

$$(4*7)^{18y}$$;

$$(2^2*7)^{18y}$$;

$$(2^2)^{18y}*7^{18y}$$;

$$2^{36y}*7^{18y}$$.

Hope it's clear.
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Math Expert
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17 Jul 2016, 10:49
Bunuel wrote:
abhilash53 wrote:
Hello,

How does the left hand side of this equation factor? ie the 28^18y

Sorry, a bit rusty on the the exponent rules. Thanks in advance.

$$28^{18y}$$;

$$(4*7)^{18y}$$;

$$(2^2*7)^{18y}$$;

$$(2^2)^{18y}*7^{18y}$$;

$$2^{36y}*7^{18y}$$.

Hope it's clear.

Theory on Exponents: math-number-theory-88376.html
Tips on Exponents: exponents-and-roots-on-the-gmat-tips-and-hints-174993.html

All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39
All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60

Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html

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Current Student
Joined: 05 Dec 2015
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07 Aug 2016, 15:40
I rewrote the equation as:
7^-x∗8^-12=28^-18y
7^-x∗2^3(-12) = 7^-18y∗ 2^2(-18y)
7^-x∗2^-36 = 7^-18y∗ 2^-36y
from here we can see that -x = -18y or x = 18y and -36=-36y or y =1.
If y is 1 then x = 18
y-x = 1-18 = -17
Intern
Joined: 30 Apr 2017
Posts: 14

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24 Jun 2017, 09:52
Bunuel
Hi Bunuel, if x & y are not integers, does it give infinitely many solutions of x & y, or does it still give finite (but different) solutions? Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 49993

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24 Jun 2017, 10:43
kerin wrote:
Bunuel
Hi Bunuel, if x & y are not integers, does it give infinitely many solutions of x & y, or does it still give finite (but different) solutions? Thanks!

It will have infinite number of solutions. For every x, there will be some y to satisfy the equation and vise-versa.
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Joined: 19 Mar 2018
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01 Aug 2018, 01:19
i did this question in following way:
simplified the given equation into 7^(18y-x) = 2^36(1-y)
now as 7 and 2 are prime no.s , LHS will be equal to RHS only when when respective powers of 7 and 2 is 0
hence, i put 18y-x=0, this implies 18y=x.
and 36(1-y)=0, this implies y=1
hence y-x= 1-18= -17

is above method correct?
Math Expert
Joined: 02 Sep 2009
Posts: 49993

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01 Aug 2018, 01:24
Kuku2206 wrote:
i did this question in following way:
simplified the given equation into 7^(18y-x) = 2^36(1-y)
now as 7 and 2 are prime no.s , LHS will be equal to RHS only when when respective powers of 7 and 2 is 0
hence, i put 18y-x=0, this implies 18y=x.
and 36(1-y)=0, this implies y=1
hence y-x= 1-18= -17

is above method correct?

_____________
Yes, it's correct.
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Re: M07-16 &nbs [#permalink] 01 Aug 2018, 01:24
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# M07-16

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