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If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)

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If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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New post 16 Jul 2014, 12:07
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If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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New post 16 Jul 2014, 12:08
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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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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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New post 16 Jul 2014, 17:53
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Bunuel wrote:


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

Kudos for a correct solution.



\((\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{7})^{18y}(\frac{1}{4})^{18y}\)

For Now, exclude \((\frac{1}{7})\) to find \(Y\)
\((\frac{1}{8})^{12}=(\frac{1}{4})^{18y}\)
\((\frac{1}{2})^{36}=(\frac{1}{2})^{36y}\)

\(36y=36\) --> \(y=1\)

\((\frac{1}{7})^x=(\frac{1}{7})^{18y}\)
\((\frac{1}{7})^x=(\frac{1}{7})^{18(1)}\)
\(x=18\)

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

Answer is B
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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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New post 16 Jul 2014, 23:42
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Focusing on denominator of \((\frac{1}{8})^{12}\) to adjust to \(\frac{1}{4}\)

\(8^{12} = 2^{(3*12)} = 2^{36} = 2^{(2*18)} = 4^{18}\)

\((\frac{1}{7})^x * (\frac{1}{4})^{18} = (\frac{1}{7})^{18y} * (\frac{1}{4})^{18y}\)

Equating powers of similar terms:

x = 18y & 18 = 18y

y = 1; x = 18

y-x = 1-18 = -17

Answer = B
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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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New post 20 Jul 2014, 04:20
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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

Sol: The given expression can be re-written as \((\frac{1}{7})^x * (\frac{1}{2^3})^{12}= (\frac{1}{7})^{18y} *(\frac{1}{2^2})^{18y}\)

Equating powers we get on both sides

x=18y
and 36=36y or y =1
x=18

y-x=-17.

Ans is B
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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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New post 20 Jul 2014, 07:44
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Bunuel wrote:


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

Kudos for a correct solution.




\(\frac{1}{7^x}*\frac{1}{8^{12}}=\frac{1}{7^{18y}}*\frac{1}{2^{36y}}\)
\(\frac{1}{7^x}*\frac{1}{2^{36}}=\frac{1}{7^{18y}}*\frac{1}{2^{36y}}\)
y = 1
x=18
y-x = -17
Answer B
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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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New post 23 Jul 2014, 10:33
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.

Kudos points given to correct solutions.

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If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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New post 04 Mar 2017, 23:23
Bunuel wrote:


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

Kudos for a correct solution.



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

Or, \((\frac{1}{7})^x*(\frac{1}{2^3})^{12}=(\frac{1}{2^2*7})^{18y}\)

Or, \((\frac{1}{7})^x*(\frac{1}{2})^{36}=(\frac{1}{2^{36y}*7^{18y}})\)

Now, \(36y = 36\)

So, \(y = 1\) and \(x = 18\)

Or, \(y - x\) = \(1 - 18\)

So, Answer will be (B) \(- 17\)
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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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New post 23 May 2018, 07:40
Bunuel wrote:
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.

Try NEW Exponents and Roots DS question.



Hi pushpitkc

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

i get \((\frac{1}{56})^{12+x}=(\frac{1}{28})^{18y}\); after this i get lost, can you please explain how Bunuel arrived at correct answer? :?

thank you :-)
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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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New post 23 May 2018, 11:28
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dave13 wrote:
Bunuel wrote:
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.

Try NEW Exponents and Roots DS question.



Hi pushpitkc

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

i get \((\frac{1}{56})^{12+x}=(\frac{1}{28})^{18y}\); after this i get lost, can you please explain how Bunuel arrived at correct answer? :?

thank you :-)


Hi dave13

The rule for exponents is \(a^x * a^y = a^{x+y}\)
However, if we have different bases, we cannot add the exponents \(a^x * b^y\) is not equal to \(ab^{x+y}\)
As a result, what you have done is wrong. Bunuel has solved this problem in the easiest possible way.

Check his solution here. If you have any doubts, please share it. I will try to help you :)
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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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New post 24 May 2018, 12:56
Abhishek009 wrote:
Bunuel wrote:


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

Kudos for a correct solution.



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

Or, \((\frac{1}{7})^x*(\frac{1}{2^3})^{12}=(\frac{1}{2^2*7})^{18y}\)

Or, \((\frac{1}{7})^x*(\frac{1}{2})^{36}=(\frac{1}{2^{36y}*7^{18y}})\)

Now, \(36y = 36\)

So, \(y = 1\) and \(x = 18\)

Or, \(y - x\) = \(1 - 18\)

So, Answer will be (B) \(- 17\)



hey pushpitkc :-)

thanks for the hint, but i stil dont get how from this \((\frac{1}{7})^x*(\frac{1}{2})^{36}=(\frac{1}{2^{36y}*7^{18y}})\)

we get this \(36y = 36\) :?
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If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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New post 24 May 2018, 19:17
1
dave13 wrote:
Abhishek009 wrote:
Bunuel wrote:


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

Kudos for a correct solution.



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

Or, \((\frac{1}{7})^x*(\frac{1}{2^3})^{12}=(\frac{1}{2^2*7})^{18y}\)

Or, \((\frac{1}{7})^x*(\frac{1}{2})^{36}=(\frac{1}{2^{36y}*7^{18y}})\)

Now, \(36y = 36\)

So, \(y = 1\) and \(x = 18\)

Or, \(y - x\) = \(1 - 18\)

So, Answer will be (B) \(- 17\)



hey pushpitkc :-)

thanks for the hint, but i stil dont get how from this \((\frac{1}{7})^x*(\frac{1}{2})^{36}=(\frac{1}{2^{36y}*7^{18y}})\)

we get this \(36y = 36\) :?


Hi dave13

We know that \(1^{anything} = 1\)

\((\frac{1}{7})^x*(\frac{1}{2})^{36}=(\frac{1}{2^{36y}*7^{18y}})\) --> \(\frac{1}{7^x}*\frac{1}{2^{36}}=(\frac{1}{2^{36y}*7^{18y}})\)

Cross-multiplying, we will get \(2^{36y}*7^{18y} = 7^x*2^{36}\)

If we have \(a^x*b^y = a^w*b^z\), x = w and y = z

Applying this learning to the above equation, we will get 36y = 36 | 18y = x

\(y = \frac{36}{36} = 1\)
\(18y = x\) -> \(x = 18\) (because y = 1)

Therefore, the value of the expression y-x is 1-18 = -17(Option B)

Hope this helps you!
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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)   [#permalink] 11 Jun 2019, 10:58
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