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 Author: Bunuel [ 16 Jul 2014, 12:07 ] Post subject: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) New project from GMAT Club: Topic-wise questions with tips and hints!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. -18B. -17C. 1D. 17E. 18

 Author: Bunuel [ 16 Jul 2014, 12:08 ] Post subject: Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) SOLUTIONIf $$x$$ 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. -18B. -17C. 1D. 17E. 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.

 Author: justbequiet [ 16 Jul 2014, 17:53 ] Post subject: Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) Bunuel wrote:New project from GMAT Club: Topic-wise questions with tips and hints!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. -18B. -17C. 1D. 17E. 18Kudos 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

 Author: PareshGmat [ 16 Jul 2014, 23:42 ] Post subject: Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) 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 = 18yy = 1; x = 18y-x = 1-18 = -17Answer = B

 Author: WoundedTiger [ 20 Jul 2014, 04:20 ] Post subject: Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) 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. -18B. -17C. 1D. 17E. 18Sol: 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 sidesx=18yand 36=36y or y =1 x=18y-x=-17.Ans is B

 Author: Game [ 20 Jul 2014, 07:44 ] Post subject: Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) Bunuel wrote:New project from GMAT Club: Topic-wise questions with tips and hints!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. -18B. -17C. 1D. 17E. 18Kudos 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 = 1x=18y-x = -17Answer B

 Author: Bunuel [ 23 Jul 2014, 10:33 ] Post subject: Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) SOLUTIONIf $$x$$ 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. -18B. -17C. 1D. 17E. 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.Try NEW Exponents and Roots DS question.

 Author: Abhishek009 [ 04 Mar 2017, 23:23 ] Post subject: Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) Bunuel wrote:New project from GMAT Club: Topic-wise questions with tips and hints!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. -18B. -17C. 1D. 17E. 18Kudos 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) $$- 17$$

 Author: dave13 [ 23 May 2018, 07:40 ] Post subject: Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) Bunuel wrote:SOLUTIONIf $$x$$ 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. -18B. -17C. 1D. 17E. 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 pushpitkcfrom 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

 Author: pushpitkc [ 23 May 2018, 11:28 ] Post subject: Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) dave13 wrote:Bunuel wrote:SOLUTIONIf $$x$$ 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. -18B. -17C. 1D. 17E. 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 pushpitkcfrom 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 dave13The 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

 Author: dave13 [ 24 May 2018, 12:56 ] Post subject: Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) Abhishek009 wrote:Bunuel wrote:New project from GMAT Club: Topic-wise questions with tips and hints!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. -18B. -17C. 1D. 17E. 18Kudos 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 $$36y = 36$$

 Author: pushpitkc [ 24 May 2018, 19:17 ] Post subject: Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) dave13 wrote:Abhishek009 wrote:Bunuel wrote:New project from GMAT Club: Topic-wise questions with tips and hints!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. -18B. -17C. 1D. 17E. 18Kudos 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 dave13We 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 = zApplying 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!

 Author: bumpbot [ 21 Nov 2022, 17:30 ] Post subject: Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) Hello from the GMAT Club BumpBot!Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

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