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Bunuel
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M07-16 [#permalink] New post  15 Sep 2014, 23:34
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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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M07-16 [#permalink] New post  15 Sep 2014, 23:34
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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

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

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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Re M07-16 [#permalink] New post  19 Jan 2015, 03:13
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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Re: M07-16 [#permalink] New post  19 Jan 2015, 03:14
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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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Re: M07-16 [#permalink] New post  26 Jun 2015, 14:20
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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Re: M07-16 [#permalink] New post  27 Jun 2015, 02:43
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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] New post  22 Dec 2015, 06:16
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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Re: M07-16 [#permalink] New post  22 Dec 2015, 06:22
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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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Re: M07-16 [#permalink] New post  17 Jul 2016, 09:43
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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Re: M07-16 [#permalink] New post  17 Jul 2016, 09:49
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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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Re: M07-16 [#permalink] New post  17 Jul 2016, 09:49
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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] New post  07 Aug 2016, 14:40
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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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Re: M07-16 [#permalink] New post  24 Jun 2017, 08:52
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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Re: M07-16 [#permalink] New post  24 Jun 2017, 09:43
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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] New post  01 Aug 2018, 00:19
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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Re: M07-16 [#permalink] New post  01 Aug 2018, 00:24
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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 [#permalink] New post  16 Nov 2019, 08:11
Bunuel wrote:
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: \(28^{18y}=7^x*8^{12}\).



You can cross multiply any fraction where the dividend is 1 by just bringing up the divisor. In this question 1/28^18y is the same as 1/1 * 28^18y/1. If you cross multiply this you should get just 28^18y. The same rule applies to the other fractions in the equation. Hope this helps!

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M07-16 [#permalink] New post  25 Feb 2020, 03:24
Without doing manipulation of exponents to solve for y and x individually:

For y-x, we know that RHS = (1/7)^18y * (1/4)^18y

LHS = RHS on (1/7) where x= 18y

y-x = y-18y = -17y; since y is > 0 and integer, we would not have an answer choice where the expression = -17(2) or -17(3) since there is only 1 right answer. Therefore, -17 is definitely the right answer with -17(1).

Answer --> (B)
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Re: M07-16 [#permalink] New post  18 Jun 2020, 12:59
Please let me know where I am going wrong?
8^-12= 4^-18y

4^-24 = 4^-18y
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Re: M07-16 [#permalink] New post  18 Jun 2020, 13:18
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nikitamaheshwari wrote:
Please let me know where I am going wrong?
8^-12= 4^-18y

4^-24 = 4^-18y


8^(-12) does not equal to 4^(-24).


8^(-12) = (2^3)^(-12) = 2^(-36), while 4^(-24) = (2^2)^(-24) = 2^(-48).
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