vjns
Bunuel
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.