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Is 7^(x + 2)/49 > 1 ?

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Joined: 02 Sep 2009
Posts: 43380
Is 7^(x + 2)/49 > 1 ? [#permalink]

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New post 16 Dec 2017, 01:00
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  25% (medium)

Question Stats:

72% (01:15) correct 28% (01:12) wrong based on 39 sessions

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Re: Is 7^(x + 2)/49 > 1 ? [#permalink]

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New post 16 Dec 2017, 05:05
Bunuel wrote:
Is \(\frac{7^{(x+2)}}{49} > 1\) ?


(1) \(7^{(x - 2)} > \frac{1}{49}\)

(2) \(7^{(x - 1)} > \frac{1}{49}\)



Slightly tricky q..

\(\frac{7^{(x+2)}}{49} > 1..........7^x>1\) MEANS --- Is x>0?

lets see the statements
(1) \(7^{(x - 2)} > \frac{1}{49}...........7^{(x-2+2)}>1...........7^x>1\)
Sufficient

(2) \(7^{(x - 1)} > \frac{1}{49}...........7^{(x-1+2)}>1.......7^{(x+1)}>0\)
so x+1>0....x>-1
if x is between -1 and 0 , ans is NO
otherwise Yes
insuff

A
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Is 7^(x + 2)/49 > 1 ? [#permalink]

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New post 16 Dec 2017, 08:10
Bunuel wrote:
Is \(\frac{7^{(x+2)}}{49} > 1\) ?


(1) \(7^{(x - 2)} > \frac{1}{49}\)

(2) \(7^{(x - 1)} > \frac{1}{49}\)


Simplify: \(\frac{7^{(x+2)}}{49} > 1 = \frac{7^{(x+2)}}{(7^2)} > 1 =7^{(x+2)-2}> 1.\)

Question: Is \(x > 0?\)

(1) \(7^{(x - 2)} > \frac{1}{49} = 7^{(x - 2)} > \frac{1}{7^2} = 7^{(x - 2)} > 7^{(-2)} = x-2>-2 = x > 0.\) Sufficient.

(2) \(7^{(x - 1)} > \frac{1}{49} = 7^{(x - 1)} > \frac{1}{7^2} = 7^{(x - 1)} > 7^{(-2)} = x-1>-2 = x-1+2 > 0 = x > -1.\) Insufficient.

(A) is the answer.
Is 7^(x + 2)/49 > 1 ?   [#permalink] 16 Dec 2017, 08:10
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Is 7^(x + 2)/49 > 1 ?

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