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If x and y are distinct positive integers, such that (2^x)(16^(x1)) =
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Updated on: 25 May 2017, 14:08
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If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ? A. 4 B. 8 C. 16 D. 32 E. 64
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Originally posted by niteshwaghray on 25 May 2017, 14:00.
Last edited by Bunuel on 25 May 2017, 14:08, edited 1 time in total.
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x1)) =
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25 May 2017, 14:16
niteshwaghray wrote: If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?
A. 4 B. 8 C. 16 D. 32 E. 64 Step 1: \((2^x)(16^{x−1})=2^{x^2}\); \((2^x)(2^{4(x−1)})=2^{x^2}\); \(2^{x+4(x−1)})=2^{x^2}\); \(x+4(x−1)=x^2\); \(x=1\) or \(x=4\) Step 2: \((3^y)(3^x)=3^{x^2−4x+5}\); \((3^{x+y}=3^{x^2−4x+5}\); \(x+y=x^2−4x+5\); \(x^25x+5=y\). Now, if \(x = 1\), then \(y = 1\) too and we are told that x and y are distinct, thus \(x = 4\) and \(y = 1\). Step 3: Therefore, \((x)^{(y+1)}=4^{1+1}=16\). Answer: C.
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x1)) =
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26 May 2017, 09:06
Bunuel wrote: niteshwaghray wrote: If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?
A. 4 B. 8 C. 16 D. 32 E. 64 Step 1: \((2^x)(16^{x−1})=2^{x^2}\); \((2^x)(2^{4(x−1)})=2^{x^2}\); \(2^{x+4(x−1)})=2^{x^2}\); \(x+4(x−1)=x^2\); \(x=1\) or \(x=4\) Step 2: \((3^y)(3^x)=3^{x^2−4x+5}\); \((3^{x+y}=3^{x^2−4x+5}\); \(x+y=x^2−4x+5\); \(x^25x+5=y\). Now, if \(x = 1\), then \(y = 1\) too and we are told that x and y are distinct, thus \(x = 4\) and \(y = 1\). Step 3: Therefore, \((x)^{(y+1)}=4^{1+1}=16\). Answer: C. Is it correct in the first step to assume that X = 4 just noticing that 1 doesn't appear in the possible answers? Thanks!



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Re: If x and y are distinct positive integers, such that (2^x)(16^(x1)) =
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26 May 2017, 09:52
MvArrow wrote: Bunuel wrote: niteshwaghray wrote: If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?
A. 4 B. 8 C. 16 D. 32 E. 64 Step 1: \((2^x)(16^{x−1})=2^{x^2}\); \((2^x)(2^{4(x−1)})=2^{x^2}\); \(2^{x+4(x−1)})=2^{x^2}\); \(x+4(x−1)=x^2\); \(x=1\) or \(x=4\) Step 2: \((3^y)(3^x)=3^{x^2−4x+5}\); \((3^{x+y}=3^{x^2−4x+5}\); \(x+y=x^2−4x+5\); \(x^25x+5=y\). Now, if \(x = 1\), then \(y = 1\) too and we are told that x and y are distinct, thus \(x = 4\) and \(y = 1\). Step 3: Therefore, \((x)^{(y+1)}=4^{1+1}=16\). Answer: C. Is it correct in the first step to assume that X = 4 just noticing that 1 doesn't appear in the possible answers? Thanks! Good observation. If x were 1, then the answer would be 1 because 1^(any number) = 1. So, yes we could say after Step 1, that x = 4.
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x1)) =
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31 May 2017, 19:14
niteshwaghray wrote: If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?
A. 4 B. 8 C. 16 D. 32 E. 64 Let’s simplify the first equation: (2^x)(16^x1) = 2^(x^2) (2^x)(2^4)^(x1) = 2^(x^2) (2^x)(2^(4x4)) = 2^(x^2) 2^(5x  4) = 2^(x^2) Because the bases are each 2, we can equate just the exponents: 5x  4 = x^2 x^2  5x + 4 = 0 (x  4)(x  1) = 0 x = 4 or x = 1 Now let’s simplify the second equation: (3^y)(3^x) = 3^(x^2  4x + 5) 3^(y + x) = 3^(x^2  4x + 5) y + x = x^2  4x + 5 y = x^2  5x + 5 Recall from our earlier work that x^2  5x + 4 = 0; thus, y = x^2  5x + 5 = (x^2  5x + 4) + 1 = 0 + 1 = 1. Furthermore, recall that x = 4 or x = 1 and we are given that x and y are distinct integers, so x must be equal to 4. Thus, x^(y + 1) = 4^(1 + 1) = 4^2 = 16. Answer: C
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x1)) =
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01 Jun 2017, 02:33
Since \((2^x)(16^{x−1})=2^{x^2}\) We can clearly say that \(x+4x4 = x^2\) because \(16 = 2^4\) Therefore, \(x^2 = 5x4\) \((3^y)(3^x)=3^{x^2−4x+5}\) Hence,\(y+x = {x^2−4x+5}\) Substituting value of x^2 in this equation y + x = 5x  4 4x + 5 y + x = x + 1 or y=1. Substituting y=1, \((x)^{(y+1)}\) must be a perfect square. From equation \(x^2 = 5x4\) we can find out the value of x to be 1 or 4. Only 4^2 is an option given in the answer options(Option C)
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If x and y are distinct positive integers, such that (2^x)(16^(x1)) =
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10 Feb 2019, 19:53
niteshwaghray wrote: If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?
A. 4 B. 8 C. 16 D. 32 E. 64 Good night everyone! Could someone please explain to me why \(2^{x^2}\) does not mean the following? (2^m)^n = 2^mn Isn't \(2^{x^2}\) the same as (2^x)^2? Thank you so much in advance!8



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If x and y are distinct positive integers, such that (2^x)(16^(x1)) =
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10 Feb 2019, 20:09
jfranciscocuencag wrote: niteshwaghray wrote: If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?
A. 4 B. 8 C. 16 D. 32 E. 64 Good night everyone! Could someone please explain to me why \(2^{x^2}\) does not mean the following? (2^m)^n = 2^mn Isn't \(2^{x^2}\) the same as (2^x)^2? Thank you so much in advance!8 Hey jfranciscocuencag(2^m)^n = 2^mn , The presence of bracket makes all the difference, As per PEMDAS, we solve the parenthesis first But if you see here, there is no parenthesis, because of that i will first solve the exponent and not the expression \(2^{x^2}\) Have a look here as well https://www.mathsisfun.com/operationorderpemdas.html
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x1)) =
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10 Feb 2019, 21:55
jfranciscocuencag wrote: niteshwaghray wrote: If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?
A. 4 B. 8 C. 16 D. 32 E. 64 Good night everyone! Could someone please explain to me why \(2^{x^2}\) does not mean the following? (2^m)^n = 2^mn Isn't \(2^{x^2}\) the same as (2^x)^2? Thank you so much in advance!8 \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\) (if exponentiation is indicated by stacked symbols, the rule is to work from the top down). Thus, \(2^{x^2}=2^{(x^2)}\) and not \((2^x)^2=2^{2x}\) On the other hand, \((a^m)^n=a^{mn}\). 8. Exponents and Roots of Numbers Check below for more: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative Megathread
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x1)) =
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