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If x and y are integers, and x ≠ 0, what is the value of x^y?

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If x and y are integers, and x ≠ 0, what is the value of x^y?  [#permalink]

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New post 02 Jun 2015, 07:03
2
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Difficulty:

  45% (medium)

Question Stats:

70% (02:07) correct 30% (02:27) wrong based on 359 sessions

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Re: If x and y are integers, and x ≠ 0, what is the value of x^y?  [#permalink]

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New post 02 Jun 2015, 07:13
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Bunuel wrote:
If x and y are integers, and x ≠ 0, what is the value of x^y?

(1) |x| = 2
(2) 64^x*6^(2x + y) = 48^(2x)


Question : what is the value of x^y?

Statement 1: |x| = 2

i.e. x = +2
but the value of y is unknown in absence of which the value of x^y is indeterminable
Hence NOT SUFFICIENT

Statement 2: 64^x*6^(2x + y) = 48^(2x)

i.e. 2^(6x) * 2^(2x+y) * 3^ (2x+y) = 2^(8x) * 3^(2x)
i.e. 2^(8x+y) * 3^ (2x+y) = 2^(8x) * 3^(2x)
Comparing the powers
i.e. (8x+y) = 8x and (2x+y) = 2x
i.e. y=0
Therefore for any value of x, the value of x^y is always 1 [Because, x^0 = 1]
Hence SUFFICIENT

Answer: Option
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Re: If x and y are integers, and x ≠ 0, what is the value of x^y?  [#permalink]

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New post 08 Jun 2015, 03:34
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Bunuel wrote:
If x and y are integers, and x ≠ 0, what is the value of x^y?

(1) |x| = 2
(2) 64^x*6^(2x + y) = 48^(2x)


MANHATTAN GMAT OFFICIAL SOLUTION:

Since x ≠ 0, we know that xy does not equal 0y or 0. To determine the exact non-zero value of xy, we need either the values of both x and y, or if y is even, the values of |x| and y (an even exponent “hides the sign” of the base, so we wouldn't need to know x's sign).

(1) INSUFFICIENT: This statement tells us that x is equal to 2 or –2. However, we know nothing about y and cannot determine the value of xy.

(2) SUFFICIENT: Simplify using exponent rules, noting the common factors of 6 and 8 on each side of the equation:

\(64^x*6^{2x + y} = 48^{2x}\)

\((8^2)^x *6^{2x + y} = (6*8)^{2x}\)

\(8^{2x}* (6^{2x}*6^y) = 6^{2x}*8^{2x}\)

\(6^y = 1\)

\(y = 0\)

Since y = 0 and x ≠ 0 (as stated in the question stem), this information is sufficient to conclude that \(x^y = x^0 = 1\).

The correct answer is B.
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Re: If x and y are integers, and x ≠ 0, what is the value of x^y?  [#permalink]

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New post 13 Mar 2017, 04:17
If x and y are integers, and x ≠ 0, what is the value of x^y?
(1) |x| = 2

(2) 64^x6^(2x + y) = 48^2x
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Re: If x and y are integers, and x ≠ 0, what is the value of x^y?  [#permalink]

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Re: If x and y are integers, and x ≠ 0, what is the value of x^y?  [#permalink]

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New post 25 Sep 2017, 18:27
Bunuel wrote:
If x and y are integers, and x ≠ 0, what is the value of \(x^y\)?

(1) |x| = 2

(2) \(64^x*6^{(2x + y)} = 48^{(2x)}\)


St 1

Too many possibilities...and no info about y

insuff

St 2

2^6x * 6^2x+y= 2^6x * 6^2x

2^6x * 6^2x+y=2^6x * 6^2x

6^(2x+y)= 6^(2x)
2x +y =2x
y= 0

This just basically tells us that x^y will always be 1 since x cannot be 0

B
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Re: If x and y are integers, and x ≠ 0, what is the value of x^y?  [#permalink]

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New post 26 Jan 2019, 04:35
Bunuel wrote:
If x and y are integers, and x ≠ 0, what is the value of \(x^y\)?

(1) |x| = 2

(2) \(64^x*6^{(2x + y)} = 48^{(2x)}\)


1) |x| = 2
Tells nothing about x

2) \(64^x*6^{(2x + y)} = 48^{(2x)}\)

\(2^{6x} * 2^{(2x + y)} * 3^{(2x + y)} = 2^{8x} * 3^{2x}\)

\(2^{(8x + y)} * 3^{(2x + y)} = 2^{8x} * 3^{2x}\)

Compare the similar terms to get value of y as 0

Now anything raised to power 0 is 1

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Re: If x and y are integers, and x ≠ 0, what is the value of x^y?  [#permalink]

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New post 24 Feb 2019, 00:45
Any tips for reducing solving time? I understand the mechanics of how to solve but can't manage to complete (2) in the requisite time.
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Re: If x and y are integers, and x ≠ 0, what is the value of x^y?  [#permalink]

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New post 24 Feb 2019, 07:43
sfoo wrote:
Any tips for reducing solving time? I understand the mechanics of how to solve but can't manage to complete (2) in the requisite time.


Hey sfoo

You have to get comfortable with playing around exponents, solve questions pertaining to this topic, this can actually help you on improving on this question type.

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Re: If x and y are integers, and x ≠ 0, what is the value of x^y?   [#permalink] 24 Feb 2019, 07:43
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