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Manager  Joined: 29 Nov 2011
Posts: 70
The expression x[n]y is defined for positive values of x and  [#permalink]

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24 00:00

Difficulty:   95% (hard)

Question Stats: 52% (03:12) correct 48% (03:00) wrong based on 292 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:

xy = 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 xy = 2, then x =

A. ¼
B. ½
C. 1
D. 2
E. 4
Math Expert V
Joined: 02 Sep 2009
Posts: 60627
Re: The expression x[n]y is defined for positive values of x and  [#permalink]

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5
7
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:
xy = 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 xy = 2, then x =

A. ¼
B. ½
C. 1
D. 2
E. 4

Given:
$$xy = x^y$$
If $$n$$ is even, $$x[n]y = (x[n-1]y)^x$$ (notice that it's the same as: if n is odd, x[n+1]y = (x[n]y)^x);
If $$n$$ is odd, $$x[n]y = (x[n-1]y)^y$$ (notice that it's the same as: if n is even, x[n+1]y = (x[n]y)^y)

Since 4 is even then $$xy=(xy)^x$$;
Since 3 is odd then $$(xy)^x=((xy)^y)^x=(xy)^{xy}$$;
Since 2 is even then $$(xy)^{xy}=((xy)^x)^{xy}=(xy)^{x^2y}$$;
Since $$xy = x^y$$ then $$(xy)^{x^2y}=(x^y)^{x^2y}=x^{x^2y^2}$$;

So, finally we have that: $$xy=x^{x^2y^2}=2$$. Now, as $$y=\frac{1}{2}$$ then $$x^{\frac{x^2}{4}}=2$$ --> raise to fourth power: $$x^{x^2}=16$$ --> $$x=2$$ or $$x=-2$$.

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Re: The expression x[n]y is defined for positive values of x and  [#permalink]

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xy = (xy)^y = ((xy)^x)^y = ((((xy)^y)^x)^y) = (xy)^(x*y^2) = (xy)^x

Now xy = 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  [#permalink]

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42% right guesses , really a tough problem Intern  Joined: 05 Jun 2014
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The expression x[n]y is defined for positive values of x and  [#permalink]

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Bunuel, how can x be = -2 ?

$$(x)^(x)^(2)$$ ----> $$-2^(-2)^2$$ = -1/16 .. Am I right?
Manager  Joined: 27 Oct 2013
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Re: The expression x[n]y is defined for positive values of x and  [#permalink]

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Well, I would say a time consuming question...

I followed top-to-bottom approach for this problem...

starting from XY, I calculated the values till XY
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)
Math Expert V
Joined: 02 Sep 2009
Posts: 60627
Re: The expression x[n]y is defined for positive values of x and  [#permalink]

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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  [#permalink]

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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.
Manager  S
Joined: 06 Oct 2015
Posts: 86
Re: The expression x[n]y is defined for positive values of x and  [#permalink]

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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:
xy = 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 xy = 2, then x =

A. ¼
B. ½
C. 1
D. 2
E. 4

Given:
$$xy = x^y$$
If $$n$$ is even, $$x[n]y = (x[n-1]y)^x$$ (notice that it's the same as: if n is odd, x[n+1]y = (x[n]y)^x);
If $$n$$ is odd, $$x[n]y = (x[n-1]y)^y$$ (notice that it's the same as: if n is even, x[n+1]y = (x[n]y)^y)

Since 4 is even then $$xy=(xy)^x$$;
Since 3 is odd then $$(xy)^x=((xy)^y)^x=(xy)^{xy}$$;
Since 2 is even then $$(xy)^{xy}=((xy)^x)^{xy}=(xy)^{x^2y}$$;
Since $$xy = x^y$$ then $$(xy)^{x^2y}=(x^y)^{x^2y}=x^{x^2y^2}$$;

So, finally we have that: $$xy=x^{x^2y^2}=2$$. Now, as $$y=\frac{1}{2}$$ then $$x^{\frac{x^2}{4}}=2$$ --> raise to fourth power: $$x^{x^2}=16$$ --> $$x=2$$ or $$x=-2$$.

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?
Math Expert V
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Posts: 60627
Re: The expression x[n]y is defined for positive values of x and  [#permalink]

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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:
xy = 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 xy = 2, then x =

A. ¼
B. ½
C. 1
D. 2
E. 4

Given:
$$xy = x^y$$
If $$n$$ is even, $$x[n]y = (x[n-1]y)^x$$ (notice that it's the same as: if n is odd, x[n+1]y = (x[n]y)^x);
If $$n$$ is odd, $$x[n]y = (x[n-1]y)^y$$ (notice that it's the same as: if n is even, x[n+1]y = (x[n]y)^y)

Since 4 is even then $$xy=(xy)^x$$;
Since 3 is odd then $$(xy)^x=((xy)^y)^x=(xy)^{xy}$$;
Since 2 is even then $$(xy)^{xy}=((xy)^x)^{xy}=(xy)^{x^2y}$$;
Since $$xy = x^y$$ then $$(xy)^{x^2y}=(x^y)^{x^2y}=x^{x^2y^2}$$;

So, finally we have that: $$xy=x^{x^2y^2}=2$$. Now, as $$y=\frac{1}{2}$$ then $$x^{\frac{x^2}{4}}=2$$ --> raise to fourth power: $$x^{x^2}=16$$ --> $$x=2$$ or $$x=-2$$.

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:
xy = 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
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Manager  S
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Posts: 86
Re: The expression x[n]y is defined for positive values of x and  [#permalink]

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Bunuel,
But you have converted x[n+1]y = (x[n]y)^x to x[n]y = (x[n-1]y)^y when n is odd and same when n is even.
Math Expert V
Joined: 02 Sep 2009
Posts: 60627
Re: The expression x[n]y is defined for positive values of x and  [#permalink]

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1
NaeemHasan wrote:
Bunuel,
But you have converted x[n+1]y = (x[n]y)^x to x[n]y = (x[n-1]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 xy=(xy)^x;

Since 3 is odd then (xy)^x=((xy)^y)^x=(xy)^(xy);

Since 2 is even then (xy)^(xy)=((xy)^x)^(xy)=(xy)^(x^2y);
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The expression x[n]y is defined for positive values of x and  [#permalink]

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Bunuel I've got doubt regarding the answer you provided.

Given:
$$xy = x^y$$
If $$n$$ is even, $$x[n]y = (x[n-1]y)^x$$ (notice that it's the same as: if n is odd, x[n+1]y = (x[n]y)^x);
If $$n$$ is odd, $$x[n]y = (x[n-1]y)^y$$ (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  [#permalink]

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2
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:

xy = 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 xy = 2, then x =

A. ¼
B. ½
C. 1
D. 2
E. 4

$$xy = (xy)^x = (xy)^{xy} = (xy)^{x^2y} = (x^y)^{x^2y} = x^{(xy)^2}$$

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

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Re: The expression x[n]y is defined for positive values of x and  [#permalink]

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_________________ Re: The expression x[n]y is defined for positive values of x and   [#permalink] 04 Dec 2019, 01:58
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