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If x ≠ 0, is a^x > y?

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Bunuel
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If x ≠ 0, is a^x > y? [#permalink] New post   29 Dec 2016, 11:47
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If x ≠ 0, is a^x > y?

(1) a = -y = | x |
(2) a < 1
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vitaliyGMAT
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Re: If x ≠ 0, is a^x > y? [#permalink] New post   29 Dec 2016, 12:59
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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.

Answer A
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AR15J
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Re: If x ≠ 0, is a^x > y? [#permalink] New post   29 Dec 2016, 14:12
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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.

Answer A



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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vitaliyGMAT
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If x ≠ 0, is a^x > y? [#permalink] New post   Updated on: 29 Dec 2016, 16:47
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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.

Answer A



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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AR15J
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Re: If x ≠ 0, is a^x > y? [#permalink] New post   29 Dec 2016, 14:43
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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 :cry:
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Re: If x ≠ 0, is a^x > y? [#permalink] New post   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.

Answer A


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] New post   07 Oct 2021, 06:41
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Re: If x ≠ 0, is a^x > y? [#permalink]
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