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If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) https://gmatclub.com/forum/ifxandyarepositiveintegersand17x174509.html 
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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) 
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 \(yx\)? A. 18 B. 17 C. 1 D. 17 E. 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) 
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 \(yx\)? A. 18 B. 17 C. 1 D. 17 E. 18 \((\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}\); Crossmultiply: \(28^{18y}=7^x*8^{12}\); 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: \(18*1=x\) > \(x=18\); \(yx=118=17\). 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: 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 \(yx\)? 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 \(Y\) \((\frac{1}{8})^{12}=(\frac{1}{4})^{18y}\) \((\frac{1}{2})^{36}=(\frac{1}{2})^{36y}\) \(36y=36\) > \(y=1\) \((\frac{1}{7})^x=(\frac{1}{7})^{18y}\) \((\frac{1}{7})^x=(\frac{1}{7})^{18(1)}\) \(x=18\) \(yx=118=17\) 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 = 18y y = 1; x = 18 yx = 118 = 17 Answer = 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 \(yx\)? A. 18 B. 17 C. 1 D. 17 E. 18 Sol: The given expression can be rewritten 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 yx=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: 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 \(yx\)? 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 yx = 17 Answer 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) 
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 \(yx\)? A. 18 B. 17 C. 1 D. 17 E. 18 \((\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}\); Crossmultiply: \(28^{18y}=7^x*8^{12}\); 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: \(18*1=x\) > \(x=18\); \(yx=118=17\). 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: 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 \(yx\)? 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) \( 17\) 
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: 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 \(yx\)? A. 18 B. 17 C. 1 D. 17 E. 18 \((\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}\); Crossmultiply: \(28^{18y}=7^x*8^{12}\); 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: \(18*1=x\) > \(x=18\); \(yx=118=17\). 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 
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: 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 \(yx\)? A. 18 B. 17 C. 1 D. 17 E. 18 \((\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}\); Crossmultiply: \(28^{18y}=7^x*8^{12}\); 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: \(18*1=x\) > \(x=18\); \(yx=118=17\). 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 
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: 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 \(yx\)? 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 \(36y = 36\) 
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: 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 \(yx\)? 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 \(1^{anything} = 1\) \((\frac{1}{7})^x*(\frac{1}{2})^{36}=(\frac{1}{2^{36y}*7^{18y}})\) > \(\frac{1}{7^x}*\frac{1}{2^{36}}=(\frac{1}{2^{36y}*7^{18y}})\) Crossmultiplying, 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 yx is 118 = 17(Option B) Hope this helps you! 
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Post subject:  Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28) 
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