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Ix x>3?

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Ix x>3? [#permalink]

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Is x>3?

1) \(5^y > 25^8\) and \(y=x^2\)
2) \(2^{15x} > 8^{4x}*8\)
[Reveal] Spoiler: OA

Last edited by ENGRTOMBA2018 on 16 Oct 2015, 05:07, edited 2 times in total.
Formatted the question, updated the OA and renamed the topic
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Re: Ix x>3? [#permalink]

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New post 14 Oct 2015, 12:33
2) 2^15x > (8^4x)(8)
Am I right that we have finally here
2^15x>2^15x ? (imagining 8 as 2^3)
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Re: Ix x>3? [#permalink]

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New post 14 Oct 2015, 22:15
Amidro wrote:
2) 2^15x > (8^4x)(8)
Am I right that we have finally here
2^15x>2^15x ? (imagining 8 as 2^3)


Here is the clarification

==>(8^4x)*(8) can be written as (2^(3*4x)) * (2^3)=2^(12x+3).
==>(2^a) *(2^b)= 2^(a+b)
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Re: Ix x>3? [#permalink]

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New post 14 Oct 2015, 23:14
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harish1986 wrote:
Is x>3?

1) \(5^y > 25^8\) and \(y=x^2\)
2) \(2^{15x} > 8^{4x}*8\)


1) \(5^y > 25^8\) and \(y=x^2\)
\(5^y > (5^2)^8\)
\(5^y > 5^{16}\)
So \(y > 16\)
\(x^2 > 16\)
\(|x| > 4\)
\(x < -4\) or \(x > 4\)
Not sufficient.


2) \(2^{15x} > 8^{4x}*8\)
\(2^{15x} > 2^{12x}*2^3\)
\(2^{15x} > 2^{12x+3}\)
15x > 12x + 3
x > 1
Not sufficient.

Both together, we know that x must be greater than 4.

Answer (C)
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Joined: 02 Oct 2015
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Location: Uzbekistan
Concentration: Entrepreneurship, Strategy
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GPA: 3.9
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Re: Ix x>3? [#permalink]

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New post 15 Oct 2015, 03:27
VeritasPrepKarishma wrote:
harish1986 wrote:
Is x>3?

1) \(5^y > 25^8\) and \(y=x^2\)
2) \(2^{15x} > 8^{4x}*8\)


1) \(5^y > 25^8\) and \(y=x^2\)
\(5^y > (5^2)^8\)
\(5^y > 5^{16}\)
So \(y > 16\)
\(x^2 > 16\)
\(|x| > 4\)
\(x < -4\) or \(x > 4\)
Not sufficient.


2) \(2^{15x} > 8^{4x}*8\)
\(2^{15x} > 2^{12x}*2^3\)
\(2^{15x} > 2^{12x+3}\)
15x > 12x + 3
x > 1
Not sufficient.

Both together, we know that x must be greater than 4.

Answer (C)

If 5^y is > than 5^16
Meaning y=17 and thus making it positive (if negative, statement 1 is not true)
Could you please clarify (y cannot equal 16, it is MORE than 16, meaning at least "17"?)
Thanks in advance!
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Re: Ix x>3? [#permalink]

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New post 15 Oct 2015, 21:03
1
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KUDOS
Expert's post
Amidro wrote:
VeritasPrepKarishma wrote:
harish1986 wrote:
Is x>3?

1) \(5^y > 25^8\) and \(y=x^2\)
2) \(2^{15x} > 8^{4x}*8\)


1) \(5^y > 25^8\) and \(y=x^2\)
\(5^y > (5^2)^8\)
\(5^y > 5^{16}\)
So \(y > 16\)
\(x^2 > 16\)
\(|x| > 4\)
\(x < -4\) or \(x > 4\)
Not sufficient.


2) \(2^{15x} > 8^{4x}*8\)
\(2^{15x} > 2^{12x}*2^3\)
\(2^{15x} > 2^{12x+3}\)
15x > 12x + 3
x > 1
Not sufficient.

Both together, we know that x must be greater than 4.

Answer (C)

If 5^y is > than 5^16
Meaning y=17 and thus making it positive (if negative, statement 1 is not true)
Could you please clarify (y cannot equal 16, it is MORE than 16, meaning at least "17"?)
Thanks in advance!


x and y needn't be integers. So y > 16 means y could be 16.2 or 16.348 or 17 or 18 or 267 etc.

y cannot be negative but x can be. y is greater than 16 means that absolute value of x is greater than 4 (because y = x^2)
If x is 5, x^2 = 25 = y (positive y greater than 16)
If x is -5, x^2 = 25 = y (positive y greater than 16)
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Re: Ix x>3? [#permalink]

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New post 29 Mar 2017, 04:41
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Re: Ix x>3?   [#permalink] 29 Mar 2017, 04:41
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