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Is y > -4?

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Joined: 01 Sep 2010
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05 Mar 2012, 06:46
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Question Stats:

41% (02:23) correct 59% (02:40) wrong based on 431 sessions

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Is y > -4?

(1) (1/7)^(4y) > (1/7)^(8y + 14)

(2) 4y^2 + 12 y < 0

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Re: Is y > -4? (1) (1/7)^(4y) > (1/7)^(8y + 14) (2) 4y^2 + 12  [#permalink]

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05 Mar 2012, 06:57
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Is y > -4?

(1) (1/7)^(4y) > (1/7)^(8y + 14) --> $$\frac{1}{7^{4y}}>\frac{1}{7^{8y+14}}$$ --> $$7^{8y+14}>7^{4y}$$ --> $$8y+14>4y$$ --> $$y>-3.5$$, hence $$y>-4$$. Sufficient.

(2) 4y^2 + 12 y < 0 --> reduce by 4 and factor out $$y$$: $$y(y+3)<0$$ --> roots are -3 and 0, "<" sign indicates that the solution lies between the roots: $$-3<y<0$$, hence $$y>-4$$. Sufficient.

Solving inequalities (to understand the reasoning for second statement):
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Re: Is y > -4? (1) (1/7)^(4y) > (1/7)^(8y + 14) (2) 4y^2 + 12  [#permalink]

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05 Mar 2012, 07:22
Bunuel wrote:
Is y > -4?

(1) (1/7)^(4y) > (1/7)^(8y + 14) --> $$\frac{1}{7^{4y}}>\frac{1}{7^{8y+14}}$$ --> $$7^{8y+14}>7^{4y}$$ --> $$8y+14>4y$$ --> $$y>-3.5$$, hence $$y>-4$$. Sufficient.

(2) 4y^2 + 12 y < 0 --> reduce by 4 and factor out $$y$$: $$y(y+3)<0$$ --> roots are -3 and 0, "<" sign indicates that the solution lies between the roots: $$-3<y<0$$, hence $$y>-4$$. Sufficient.

Solving inequalities (to understand the reasoning for second statement):
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

For 1) is the same (of course is subjective) to do: 7^-4y > 7^-8y+14.......in the end: y>7/2 ?????

For 2) divide by 4 we have : y(y+3)<0 the sign is LESS so, as consequence the value of x is between (not at the extreme as in case of sign > ) -3<x<0. In this case of course our variable is y but is the same, is only one unknown.

By the way OA is D but I have never seen a post from you wrong. Precious work.
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Posts: 54544
Re: Is y > -4? (1) (1/7)^(4y) > (1/7)^(8y + 14) (2) 4y^2 + 12  [#permalink]

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05 Mar 2012, 07:28
carcass wrote:
Bunuel wrote:
Is y > -4?

(1) (1/7)^(4y) > (1/7)^(8y + 14) --> $$\frac{1}{7^{4y}}>\frac{1}{7^{8y+14}}$$ --> $$7^{8y+14}>7^{4y}$$ --> $$8y+14>4y$$ --> $$y>-3.5$$, hence $$y>-4$$. Sufficient.

(2) 4y^2 + 12 y < 0 --> reduce by 4 and factor out $$y$$: $$y(y+3)<0$$ --> roots are -3 and 0, "<" sign indicates that the solution lies between the roots: $$-3<y<0$$, hence $$y>-4$$. Sufficient.

Solving inequalities (to understand the reasoning for second statement):
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

For 1) is the same (of course is subjective) to do: 7^-4y > 7^-8y+14.......in the end: y>7/2 ?????

For 2) divide by 4 we have : y(y+3)<0 the sign is LESS so, as consequence the value of x is between (not at the extreme as in case of sign > ) -3<x<0. In this case of course our variable is y but is the same, is only one unknown.

By the way OA is D but I have never seen a post from you wrong. Precious work.

The red part should be 7^(-4y) > 7^(-8y-14). Else is correct.
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Re: Is y > -4?  [#permalink]

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17 Jun 2013, 05:03
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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Re: Is y > -4? (1) (1/7)^(4y) > (1/7)^(8y + 14) (2) 4y^2 + 12  [#permalink]

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29 Oct 2013, 00:15
Bunuel wrote:
Is y > -4?
(1) (1/7)^(4y) > (1/7)^(8y + 14) --> $$\frac{1}{7^{4y}}>\frac{1}{7^{8y+14}}$$ --> $$7^{8y+14}>7^{4y}$$ --> $$8y+14>4y$$ --> $$y>-3.5$$, hence $$y>-4$$. Sufficient.

Hi Bunuel! I have read the theory for solving inequalities, but still I am not very comfortable with the subject. Could you please explain why while solving statement 1 we do not consider the sign for y? Generally while solving inequality for a variable we consider the cases when y>0 and y<0. Then in this particular case why the sign of y is not necessary? TIA.
Math Expert
Joined: 02 Sep 2009
Posts: 54544
Re: Is y > -4? (1) (1/7)^(4y) > (1/7)^(8y + 14) (2) 4y^2 + 12  [#permalink]

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29 Oct 2013, 02:36
1
vjns wrote:
Bunuel wrote:
Is y > -4?
(1) (1/7)^(4y) > (1/7)^(8y + 14) --> $$\frac{1}{7^{4y}}>\frac{1}{7^{8y+14}}$$ --> $$7^{8y+14}>7^{4y}$$ --> $$8y+14>4y$$ --> $$y>-3.5$$, hence $$y>-4$$. Sufficient.

Hi Bunuel! I have read the theory for solving inequalities, but still I am not very comfortable with the subject. Could you please explain why while solving statement 1 we do not consider the sign for y? Generally while solving inequality for a variable we consider the cases when y>0 and y<0. Then in this particular case why the sign of y is not necessary? TIA.

The sign of y has nothing to do when solving $$7^{8y+14}>7^{4y}$$.

Consider another way. Since $$7^{4y}$$ is positive irrespective of value of y, then we can safely reduce both sides by it and write: $$\frac{7^{8y+14}}{7^{4y}}>1$$ --> $$7^{8y+14-4y}>1$$.

$$7^{8y+14-4y}>1$$ to hold true the exponent must be positive: $$8y+14-4y>0$$ --> $$y>-3.5$$.

Hope this helps.
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Is y> -4? (1) (1/7)^(4y) > (1/7)^(8y+14), (2) 4y^2 + 12y < 0  [#permalink]

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24 Jul 2017, 09:43
Is $$y > -4?$$
(1) ($$\frac{1}{7})^{4y} > (\frac{1}{7})^{8y + 14}$$
(2) $$4y^2 + 12y < 0$$

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Re: Is y> -4? (1) (1/7)^(4y) > (1/7)^(8y+14), (2) 4y^2 + 12y < 0  [#permalink]

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24 Jul 2017, 10:03
Good question. Just need some careful calculations.

S1:

(1/7)^4y >(1/7)^(8y+14)

Now this can not be directly equated. You will need to write it like this.

(7)^-4y>(7)^-(8y+14)

or

-4y>-(8y+14)

hence y>-3.5

Sufficient

S2:

4y^2 +12y <0

This will give you -3<y<0

hence y >-4

Sufficient

Re: Is y> -4? (1) (1/7)^(4y) > (1/7)^(8y+14), (2) 4y^2 + 12y < 0   [#permalink] 24 Jul 2017, 10:03
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