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The expression x[n]y is defined for positive values of x and
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09 May 2012, 04:43
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52% (03:12) correct 48% (03:01) wrong based on 379 sessions
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The expression x[n]y is defined for positive values of x and y and for positive integer values of n as follows: x[1]y = x^y If n is odd, x[n+1]y = (x[n]y)^x If n is even, x[n+1]y = (x[n]y)^y If y = ½ and x[4]y = 2, then x = A. ¼ B. ½ C. 1 D. 2 E. 4
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Re: The expression x[n]y is defined for positive values of x and
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09 May 2012, 06:50
Smita04 wrote: The expression x[n]y is defined for positive values of x and y and for positive integer values of n as follows: x[1]y = x^y If n is odd, x[n+1]y = (x[n]y)^x If n is even, x[n+1]y = (x[n]y)^y
If y = ½ and x[4]y = 2, then x =
A. ¼ B. ½ C. 1 D. 2 E. 4 Given: \(x[1]y = x^y\) If \(n\) is even, \(x[n]y = (x[n1]y)^x\) ( notice that it's the same as: if n is odd, x[n+1]y = (x[n]y)^x); If \(n\) is odd, \(x[n]y = (x[n1]y)^y\) ( notice that it's the same as: if n is even, x[n+1]y = (x[n]y)^y) Since 4 is even then \(x[4]y=(x[3]y)^x\); Since 3 is odd then \((x[3]y)^x=((x[2]y)^y)^x=(x[2]y)^{xy}\); Since 2 is even then \((x[2]y)^{xy}=((x[1]y)^x)^{xy}=(x[1]y)^{x^2y}\); Since \(x[1]y = x^y\) then \((x[1]y)^{x^2y}=(x^y)^{x^2y}=x^{x^2y^2}\); So, finally we have that: \(x[4]y=x^{x^2y^2}=2\). Now, as \(y=\frac{1}{2}\) then \(x^{\frac{x^2}{4}}=2\) > raise to fourth power: \(x^{x^2}=16\) > \(x=2\) or \(x=2\). Answer: D.
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Re: The expression x[n]y is defined for positive values of x and
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09 May 2012, 05:51
x[4]y = (x[3]y)^y = ((x[2]y)^x)^y = ((((x[1]y)^y)^x)^y) = (x[1]y)^(x*y^2) = (x[1]y)^x Now x[1]y = x^(1/2) => (x^(1/2))^x = 2 => x^(x/2) = 2 => x^x = 4 => x = 2 Option D
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Re: The expression x[n]y is defined for positive values of x and
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11 Aug 2013, 04:50
42% right guesses , really a tough problem
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The expression x[n]y is defined for positive values of x and
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24 Jan 2015, 22:52
Bunuel, how can x be = 2 ?
\((x)^(x)^(2)\) > \(2^(2)^2\) = 1/16 .. Am I right?



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Re: The expression x[n]y is defined for positive values of x and
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24 Jan 2015, 23:08
Well, I would say a time consuming question...
I followed toptobottom approach for this problem...
starting from X[1]Y, I calculated the values till X[4]Y then substituted the given parameters...
Option D is correct..
Time taken ~ 3 minutes(I am very susceptible to silly mistakes, so I have to be double sure)



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Re: The expression x[n]y is defined for positive values of x and
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25 Jan 2015, 03:43
saleem1992 wrote: Bunuel, how can x be = 2 ?
\((x)^(x)^(2)\) > \(2^(2)^2\) = 1/16 .. Am I right? No. (2)^(2)^2 = (2)^4 = 16. If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^{m^n}=a^{(m^n)}\) and not \({(a^m)}^n\), which on the other hand equals to \(a^{mn}\).
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Re: The expression x[n]y is defined for positive values of x and
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23 May 2016, 01:43
Bunuel, Could you please explain by using the information given rather then changing the odd even expressions. I used the info given rather manipulating it. I am not getting the result.



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Re: The expression x[n]y is defined for positive values of x and
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14 Jun 2017, 02:35
Bunuel wrote: Smita04 wrote: The expression x[n]y is defined for positive values of x and y and for positive integer values of n as follows: x[1]y = x^y If n is odd, x[n+1]y = (x[n]y)^x If n is even, x[n+1]y = (x[n]y)^y
If y = ½ and x[4]y = 2, then x =
A. ¼ B. ½ C. 1 D. 2 E. 4 Given: \(x[1]y = x^y\) If \(n\) is even, \(x[n]y = (x[n1]y)^x\) (notice that it's the same as: if n is odd, x[n+1]y = (x[n]y)^x); If \(n\) is odd, \(x[n]y = (x[n1]y)^y\) (notice that it's the same as: if n is even, x[n+1]y = (x[n]y)^y)Since 4 is even then \(x[4]y=(x[3]y)^x\); Since 3 is odd then \((x[3]y)^x=((x[2]y)^y)^x=(x[2]y)^{xy}\); Since 2 is even then \((x[2]y)^{xy}=((x[1]y)^x)^{xy}=(x[1]y)^{x^2y}\); Since \(x[1]y = x^y\) then \((x[1]y)^{x^2y}=(x^y)^{x^2y}=x^{x^2y^2}\); So, finally we have that: \(x[4]y=x^{x^2y^2}=2\). Now, as \(y=\frac{1}{2}\) then \(x^{\frac{x^2}{4}}=2\) > raise to fourth power: \(x^{x^2}=16\) > \(x=2\) or \(x=2\). Answer: D. Hi, Bunuel! I am unable to understand the colored line. How does power shift from Y to X in the first line and X to Y in the second line?



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Re: The expression x[n]y is defined for positive values of x and
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14 Jun 2017, 02:49
NaeemHasan wrote: Bunuel wrote: Smita04 wrote: The expression x[n]y is defined for positive values of x and y and for positive integer values of n as follows: x[1]y = x^y If n is odd, x[n+1]y = (x[n]y)^x If n is even, x[n+1]y = (x[n]y)^y
If y = ½ and x[4]y = 2, then x =
A. ¼ B. ½ C. 1 D. 2 E. 4 Given: \(x[1]y = x^y\) If \(n\) is even, \(x[n]y = (x[n1]y)^x\) (notice that it's the same as: if n is odd, x[n+1]y = (x[n]y)^x); If \(n\) is odd, \(x[n]y = (x[n1]y)^y\) (notice that it's the same as: if n is even, x[n+1]y = (x[n]y)^y)Since 4 is even then \(x[4]y=(x[3]y)^x\); Since 3 is odd then \((x[3]y)^x=((x[2]y)^y)^x=(x[2]y)^{xy}\); Since 2 is even then \((x[2]y)^{xy}=((x[1]y)^x)^{xy}=(x[1]y)^{x^2y}\); Since \(x[1]y = x^y\) then \((x[1]y)^{x^2y}=(x^y)^{x^2y}=x^{x^2y^2}\); So, finally we have that: \(x[4]y=x^{x^2y^2}=2\). Now, as \(y=\frac{1}{2}\) then \(x^{\frac{x^2}{4}}=2\) > raise to fourth power: \(x^{x^2}=16\) > \(x=2\) or \(x=2\). Answer: D. Hi, Bunuel! I am unable to understand the colored line. How does power shift from Y to X in the first line and X to Y in the second line? This is given in the definition of the function: x[n]y is defined for positive values of x and y and for positive integer values of n as follows: x[1]y = x^y If n is odd, x[n+1]y = (x[n]y)^ xIf n is even, x[n+1]y = (x[n]y)^ y
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Re: The expression x[n]y is defined for positive values of x and
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14 Jun 2017, 05:11
Bunuel, But you have converted x[n+1]y = (x[n]y)^x to x[n]y = (x[n1]y)^y when n is odd and same when n is even.



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Re: The expression x[n]y is defined for positive values of x and
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14 Jun 2017, 05:35
NaeemHasan wrote: Bunuel, But you have converted x[n+1]y = (x[n]y)^x to x[n]y = (x[n1]y)^y when n is odd and same when n is even. Where did I do that? Maybe you are confused by double exponents? Since 4 is even then x[4]y=(x[3]y)^ x; Since 3 is odd then (x[3]y)^x=((x[2]y)^ y)^x=(x[2]y)^(xy); Since 2 is even then (x[2]y)^(xy)=((x[1]y)^ x)^(xy)=(x[1]y)^(x^2y);
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The expression x[n]y is defined for positive values of x and
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18 Oct 2018, 01:00
Bunuel I've got doubt regarding the answer you provided. Given: \(x[1]y = x^y\) If \(n\) is even, \(x[n]y = (x[n1]y)^x\) (notice that it's the same as: if n is odd, x[n+1]y = (x[n]y)^x); If \(n\) is odd, \(x[n]y = (x[n1]y)^y\) (notice that it's the same as: if n is even, x[n+1]y = (x[n]y)^y)how did you change the equation when the question stem states : If n is odd, x[n+1]y = (x[n]y)^x If n is even, x[n+1]y = (x[n]y)^y is there any other easy way to solve this question as it seems to be really difficult! Please help. Thanks a lot in advance
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Re: The expression x[n]y is defined for positive values of x and
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18 Oct 2018, 02:15
Smita04 wrote: The expression x[n]y is defined for positive values of x and y and for positive integer values of n as follows:
x[1]y = x^y If n is odd, x[n+1]y = (x[n]y)^x If n is even, x[n+1]y = (x[n]y)^y
If y = ½ and x[4]y = 2, then x =
A. ¼ B. ½ C. 1 D. 2 E. 4 \(x[4]y = (x[3]y)^x = (x[2]y)^{xy} = (x[1]y)^{x^2y} = (x^y)^{x^2y} = x^{(xy)^2}\) Since y = 1/2, we get \(2 = x^{(x/2)^2}\) Look for a value of x from the options which will satisfy this equation. Since the power has x/2 but LHS is an integer, x should be a multiple of 2. Try x = 2, you get \(2 = 2^{(2/2)^2} = 2^1\) Satisfies Answer (D)
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Re: The expression x[n]y is defined for positive values of x and
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