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M07-16

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New post 16 Sep 2014, 00:34
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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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New post 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: \(18*1=x\) --> \(x=18\);

\(y-x=1-18=-17\).

Answer: B.
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New post 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.
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New post 19 Jan 2015, 04:14
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New post 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.
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New post 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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Re: M07-16  [#permalink]

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New post 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.
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New post 22 Dec 2015, 07:22
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New post 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.
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New post 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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New post 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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Re: M07-16  [#permalink]

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New post 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
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New post 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!
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New post 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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Re: M07-16  [#permalink]

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New post 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?
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New post 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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