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If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:
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10 May 2012, 08:04
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If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals: A. (1,2) B. (2,1) C. (1,1) D. (2,2) E. (1,3) I understand how to cancel out things for the first half of the problem. For the second half, GMATPrep v2 says that (3^2x)(3^y) = 3^4 becomes 32x + y = 34. Can somebody explain how that happens?
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Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:
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10 May 2012, 08:09
RyanP wrote: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:
A. (1,2) B. (2,1) C. (1,1) D. (2,2) E. (1,3)
I understand how to cancel out things for the first half of the problem. For the second half, GMATPrep v2 says that (3^2x)(3^y) = 3^4 becomes 32x + y = 34. Can somebody explain how that happens? \(2^x*2^y=8\) > \(2^{x+y}=2^3\) > \(x+y=3\); \(9^x*3^y=81\) > \(3^{2x}*3^y=3^4\) > \(3^{2x+y}=3^4\) > \(2x+y=4\); Solving: \(x+y=3\) and \(2x+y=4\) we get \(x=1\) and \(y=2\). Answer: A. Hope it's clear.
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Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:
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11 May 2012, 10:03
you could simply plug in answer choices into the equations and see which one is equal... answer A.



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Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:
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28 May 2013, 16:44
If (2 to the power of X)(2 to the power of Y) and (9 to the power of X) (3 to the power of Y) = 81 then (X,Y) =
(1,2)
(2,1)
(1,1)
(2,2)
(1,3)



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Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:
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28 May 2013, 16:48



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Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:
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29 May 2013, 02:52
clicker wrote: you could simply plug in answer choices into the equations and see which one is equal... answer A. I think the plugin method works best in the actual exam when you're fighting against the clock, or when you've exhausted all possible angles of attack on a notoriously difficult problem. But during preparation, I believe it is essential to understand how and why the concept work to solve these questions. To solve this question you should be familiar with the following rules :  \(a^x * a^y = a^(x+y)\) (1) (notice that for this rule to work, we must have the same base, as in "a")  \((a^x)^y = a^(x*y)\) (2)  \(a^x = a^y\) therefore \(x = y\) (3) (same as with rule n°1, we need to have the same base) Considering our problem, we have : \(2^x*2^y = 8\) which, according to (1) will yield \(2^(x+y) = 8 = 2^3\) which, according to (3) will yield : \(x + y = 3\) \(9^x*3^y = 81\) Since \(9 = 3^2\), then the above equation becomes : \((3^2)^x*3^y = 81\). According to (2), we can write : \(3^(2*x)*3^y = 81\) which, according to (1) will yield \(3^(2x+y) = 81 = 3^4\), which according to (3), will yield : \(2x + y = 4\) We end up with a linear system of two equations : \(x + y = 3\) \(2x + y = 4\) The solution of which is x = 1 and y = 2. Which is answer choice A. Hope that helped



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Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:
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16 Oct 2016, 11:34
RyanP wrote: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals: A. (1,2) B. (2,1) C. (1,1) D. (2,2) E. (1,3) I understand how to cancel out things for the first half of the problem. For the second half, GMATPrep v2 says that (3^2x)(3^y) = 3^4 becomes 32x + y = 34. Can somebody explain how that happens? \((2^x) (2^y) = 8\) Or, \(2^{ x + y } = 2^3\) So, \(x + y = 3\)>(I) \((9^x)(3^y) = 81\) Or, \(3^{2x + y } =3^4\) So, \(2x + y = 4\)>(II) Combine (I) and (II) \(2x + 2y = 6\) \(2x + y = 4\) Solve for \(y\) \(y = 2\) Plug in y = 2 in ( I) \(x + y = 3\) Or, \(x + 2 = 3\) So, \(x = 1\) Hence (x , y ) = ( 1 , 2 ) Thus answer will be (A) (1,2)
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Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:
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22 Jun 2017, 06:54
If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals: If base is same, we add powers. So, x+y= 3 (3 because 2^3=8 and we are comparing/equating powers) 2x+y= 4 Let's subtract the two equations: 2x+y=4 xy=3 x= 1 and thus y=2 A. (1,2)B. (2,1) C. (1,1) D. (2,2) E. (1,3)
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Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:
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