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.

Answer: D.

Solving inequalities (to understand the reasoning for second statement):

x2-4x-94661.html#p731476 (

check this one first)

inequalities-trick-91482.htmldata-suff-inequalities-109078.htmlrange-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535everything-is-less-than-zero-108884.html?hilit=extreme#p868863Hope 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.

Thanks for links, I have already read; very useful .

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.