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The operation is defined for all non-zero x and y by the equation x

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The operation is defined for all non-zero x and y by the equation x [#permalink]

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New post 19 May 2017, 12:04
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The operation @ is defined for all non-zero x and y by the equation x@ y = x^y . Then the expression (x@ y)@ z is equal to

(A) x^y^z
(B) xyz
(C) (xy)^z
(D) (x^y) z
(E) (x^y)^z

Source: Nova GMAT
[Reveal] Spoiler: OA

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Re: The operation is defined for all non-zero x and y by the equation x [#permalink]

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New post 19 May 2017, 13:15
SajjadAhmad wrote:
The operation @ is defined for all non-zero x and y by the equation x@ y = x^y . Then the expression (x@ y)@ z is equal to

(A) x^y^z
(B) xyz
(C) (xy)^z
(D) (x^y) z
(E) (x^y)^z

Source: Nova GMAT


Sorry for the extra brackets. The "at" symbol used makes formatting hard; I can't figure out how to make it stop turning all that follows the symbol into a URL.

For [(x @ y) @ z], start with parentheses.

x @ y = \(x^y\), which is the new LHS ==>

[\(x^y\) @ z] is \(x^y\) to the \({z}^th\) power, which is

[Reveal] Spoiler:
(x^y)^z, Answer E.

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Re: The operation is defined for all non-zero x and y by the equation x [#permalink]

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New post 19 May 2017, 21:46
I will try to solve this using smart numbers. I will avoid using values of 1, -1, 0 as more than one options are likely to yield the same answers if I take such values here.

lets take x=2, y=3, z=2.
Now x@y = 2^3 =8
and (x@y)@z = 8@2 = 8^2 = 64

Now we have to look at an option which gives us 64.

1) x^y^z = 2^3^2 = 2^9 =512
2) xyz = 2*3*2 =12
3) (xy)^z = (2*3)^2 = 6^2 =36
4) (x^y)z = (2^3)*2 = 8*2 =16
5) (x^y)^z = (2^3)^2 = 8^2 = 64

Hence answer is E
Re: The operation is defined for all non-zero x and y by the equation x   [#permalink] 19 May 2017, 21:46
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