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Is 7^x > 100 ?

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Is 7^x > 100 ? [#permalink]

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Difficulty:

  75% (hard)

Question Stats:

52% (01:23) correct 48% (01:35) wrong based on 115 sessions

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Is 7^x > 100 ?

(1) 7^(x+2) > 9800
(2) 7^(2x) > 100,00
[Reveal] Spoiler: OA

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Re: Is 7^x > 100 ? [#permalink]

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New post 12 May 2010, 20:15
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1. 7^(x+2)> 9800

or 7^x * 7^2 > 9800
or 7^x > 9800 / 49
or 7^x > 200 ------> 7^x > 100

Therefore , sufficient.

2. 7^2X > 100,00

or (7^x)^2 > 10000
or (7^x)^2 - (100)^2 > 0 ----> a^2 - b^2 >0
implies 7^x > 100 or 7^x <-100

Therefore not sufficient.

Ans : A

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Re: Is 7^x > 100 ? [#permalink]

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harikattamudi wrote:
Is 7^X > 100 ?

(1) 7^(X+2) > 9800
(2) 7^2X > 100,00


Thanks
-H

Is it D?
(1): suff
(2): (7^x)^2>100,00 so 7^x>100. 7^x cannot be negative

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Re: Is 7^x > 100 ? [#permalink]

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New post 13 May 2010, 05:45
"7^x <-100"

ok this is not possibt because 7^x can not be negative

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Re: Is 7^x > 100 ? [#permalink]

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New post 13 May 2010, 23:59
yes agree.. thanks for correcting...

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Re: Is 7^x > 100 ? [#permalink]

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New post 15 May 2010, 08:34
7^x cannot be -ive hence the ans is D

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Re: Is 7^x > 100 ? [#permalink]

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New post 12 Nov 2016, 00:01
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harikattamudi wrote:
Is 7^X > 100 ?

(1) 7^(X+2) > 9800
(2) 7^2X > 100,00


Thanks
-H


Answer: Option D

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Re: Is 7^x > 100 ? [#permalink]

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New post 04 Dec 2016, 22:02
Can you please explain why 7^x can not be negative?

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Re: Is 7^x > 100 ? [#permalink]

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New post 09 Dec 2016, 04:01
AnotherGmater wrote:
Can you please explain why 7^x can not be negative?


if x is positive, e.g. 1 then 7^1 = 7
if x is 0 then 7^0 = 1
if x is negative then 7^-1 = 1/7

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Re: Is 7^x > 100 ? [#permalink]

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New post 09 Dec 2016, 14:16
X does not matter, but what about 7? 7 can be negative.

(-7)^x * (-7)^x = (-7)^2x since anything power 2 is positive we can write it as 7^2x.

so, when we remove square, the value can be either (-7)^x or 7^x. In that case, B is not really sufficient by itself.

Correct me if I'm wrong.

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Is 7^x > 100 ? [#permalink]

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New post 09 Dec 2016, 14:24
Bunuel can you please help me understand why B is also sufficient by itself?
By decrypting the second condition I will get (7^x-100) >0 and/or (7^x +100) >0
that means 7^x>100 which is fine, but don't we also need to take another equation into the account i.e 7^x > -100. So, from the second equation we are not sure about 7^x, since 7^x can be anywhere starting from -99.999.

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Re: Is 7^x > 100 ? [#permalink]

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New post 09 Dec 2016, 18:53
AnotherGmater wrote:
Bunuel can you please help me understand why B is also sufficient by itself?
By decrypting the second condition I will get (7^x-100) >0 and/or (7^x +100) >0
that means 7^x>100 which is fine, but don't we also need to take another equation into the account i.e 7^x > -100. So, from the second equation we are not sure about 7^x, since 7^x can be anywhere starting from -99.999.



AnotherGmater

When the question says 7^x it means +ve integer 7 has powered to x
where x can be any number and not only x=2y..

so you have to consider only 7^x>0 in any case

hope it helps...

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Re: Is 7^x > 100 ?   [#permalink] 09 Dec 2016, 18:53
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