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M07-16

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Bunuel
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M07-16 [#permalink] 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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B
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Re M07-16 [#permalink] New post   16 Sep 2014, 00:34
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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. \(18\)


Given that \((\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}\), we can rewrite this as:

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

Next, we cross-multiply:

\(28^{18y}=7^x*8^{12}\).

Now, we factorize:

\(2^{36y}*7^{18y}=7^x*2^{36}\).

To find the value of \(y\), we equate the powers of 2 on both sides:

\(36y=36\), hence \(y=1\).

To find the value of \(x\), we equate the powers of 7 on both sides:

\(18*1=x\), hence \(x=18\);

Finally, we calculate \(y-x=1-18=-17\).


Answer: B
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Re: M07-16 [#permalink] New post   17 Jul 2016, 10:49
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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, 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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kerin
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Re: M07-16 [#permalink] 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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Re: M07-16 [#permalink] New post   24 Jun 2017, 10:43
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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   18 Jun 2020, 13:59
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, 14:18
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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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Re: M07-16 [#permalink] New post   18 Apr 2023, 14:55
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Go650GMAT
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Re M07-16 [#permalink] New post   26 Mar 2024, 13:40
I think this is a high-quality question and I agree with explanation.
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Re M07-16 [#permalink]
26 Mar 2024, 13:40
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