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# If x ≠ 0, is a^x > y?

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Math Expert
Joined: 02 Sep 2009
Posts: 56366
If x ≠ 0, is a^x > y?  [#permalink]

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29 Dec 2016, 11:47
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Difficulty:

45% (medium)

Question Stats:

61% (01:43) correct 39% (01:59) wrong based on 164 sessions

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If x ≠ 0, is a^x > y?

(1) a = -y = | x |
(2) a < 1

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Joined: 13 Oct 2016
Posts: 363
GPA: 3.98
Re: If x ≠ 0, is a^x > y?  [#permalink]

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29 Dec 2016, 12:59
2
3
Bunuel wrote:
If x ≠ 0, is a^x > y?

(1) a = -y = | x |
(2) a < 1

(1) a = -y = | x |

a>0 , y<0 x>0 or x<0

x>0 $$a^x > y$$

x<0 $$\frac{1}{a^x} > y$$ Sufficient.

(2) a < 1

a<0 or 0<a<1 Insufficient.

##### General Discussion
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Re: If x ≠ 0, is a^x > y?  [#permalink]

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29 Dec 2016, 14:12
1
1
vitaliyGMAT wrote:
Bunuel wrote:
If x ≠ 0, is a^x > y?

(1) a = -y = | x |
(2) a < 1

(1) a = -y = | x |

a>0 , y<0 x>0 or x<0

x>0 $$a^x > y$$

x<0 $$\frac{1}{a^x} > y$$ Sufficient.

(2) a < 1

a<0 or 0<a<1 Insufficient.

In statement 1 , why did you neglect the condition when a<0 ? I think that the answer should be C.

Correct me if I am wrong.

+1 Kudos if you like the post
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Posts: 363
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If x ≠ 0, is a^x > y?  [#permalink]

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Updated on: 29 Dec 2016, 16:47
1
AR15J wrote:
vitaliyGMAT wrote:
Bunuel wrote:
If x ≠ 0, is a^x > y?

(1) a = -y = | x |
(2) a < 1

(1) a = -y = | x |

a>0 , y<0 x>0 or x<0

x>0 $$a^x > y$$

x<0 $$\frac{1}{a^x} > y$$ Sufficient.

(2) a < 1

a<0 or 0<a<1 Insufficient.

In statement 1 , why did you neglect the condition when a<0 ? I think that the answer should be C.

Correct me if I am wrong.

+1 Kudos if you like the post

Hi

because a = |x|, x ≠ 0 and absolute value is always non negative. From this a>0 and -y>0 ---> y<0. But our x can be either <0 or >0

Hope this makes sense.

Originally posted by vitaliyGMAT on 29 Dec 2016, 14:30.
Last edited by vitaliyGMAT on 29 Dec 2016, 16:47, edited 1 time in total.
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Re: If x ≠ 0, is a^x > y?  [#permalink]

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29 Dec 2016, 14:43
1
In statement 1 , why did you neglect the condition when a<0 ? I think that the answer should be C.

Correct me if I am wrong.

+1 Kudos if you like the post [/quote]

Hi

because a = |x| and absolute value is always positive. From this a>0 and -y>0 ---> y<0. But our x can be either <0 or >0

Hope this makes sense.[/quote]

Yes, it does. Sorry , It's too bad that I could not see it
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Re: If x ≠ 0, is a^x > y?  [#permalink]

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04 Jan 2017, 08:18
vitaliyGMAT wrote:
Bunuel wrote:
If x ≠ 0, is a^x > y?

(1) a = -y = | x |
(2) a < 1

(1) a = -y = | x |

a>0 , y<0 x>0 or x<0

x>0 $$a^x > y$$

x<0 $$\frac{1}{a^x} > y$$ Sufficient.

(2) a < 1

a<0 or 0<a<1 Insufficient.

Good explanation. I made the same mistake as AR15J, but now its clear. Thank you!
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Re: If x ≠ 0, is a^x > y?  [#permalink]

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21 Aug 2018, 07:55
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Re: If x ≠ 0, is a^x > y?   [#permalink] 21 Aug 2018, 07:55
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