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Question Number 100409 by Algoritm last updated on 26/Jun/20

Commented by bramlex last updated on 26/Jun/20
$(((3/2)×4^x )/(4^x −9^x )) > 3+ ((4/9))^x ((3.((4/9))^x )/(2{((4/9))^x −1})) > 3+((4/9))^x let ((4/9))^x = z > 0 _(1) ((3z)/(2(z−1))) > 3+z ⇒((3z−(3+z)(z−1))/(2(z−1))) > 0 ((3z−(z^2 +2z−3))/(z−1)) > 0 ((z^2 −z−3)/(z−1)) < 0 ⇒(((z−(1/2))^2 −((13)/4))/(z−1)) < 0 (((z−(1/2)−((√(13))/2))(z−(1/2)+((√(13))/2)))/(z−1)) < 0 z < ((1−(√(13)))/2) or 1< z < ((1+(√(13)))/2) _(2) solution for z : (1)∩(2) 1 < z < ((1+(√(13)))/2) ⇒ 1< ((4/9))^x <((1+(√(13)))/2) log _(((4/9))) (((1+(√(13)))/2)) < x < 0 −1.0285 < x < 0$
$\frac{\frac{3}{2} \times 4^{x}}{4^{x} - 9^{x}} > 3 + {(\frac{4}{9})}^{x}$ $\frac{3 . {(\frac{4}{9})}^{x}}{2 {{(\frac{4}{9})}^{x} - 1}} > 3 + {(\frac{4}{9})}^{x}$ $l e t {(\frac{4}{9})}^{x} = z > 0_(1)$ $\frac{3 z}{2 (z - 1)} > 3 + z \Rightarrow \frac{3 z - (3 + z) (z - 1)}{2 (z - 1)} > 0$ $\frac{3 z - (z^{2} + 2 z - 3)}{z - 1} > 0$ $\frac{z^{2} - z - 3}{z - 1} < 0 \Rightarrow \frac{{(z - \frac{1}{2})}^{2} - \frac{13}{4}}{z - 1} < 0$ $\frac{(z - \frac{1}{2} - \frac{\sqrt{13}}{2}) (z - \frac{1}{2} + \frac{\sqrt{13}}{2})}{z - 1} < 0$ $z < \frac{1 - \sqrt{13}}{2} o r 1 < z < \frac{1 + \sqrt{13}}{2}_(2)$ $s o l u t i o n f o r z : (1) \cap (2)$ $1 < z < \frac{1 + \sqrt{13}}{2} \Rightarrow 1 < {(\frac{4}{9})}^{x} < \frac{1 + \sqrt{13}}{2}$ $\log_{(\frac{4}{9})} (\frac{1 + \sqrt{13}}{2}) < x < 0$ $- 1.0285 < x < 0$
Commented by Algoritm last updated on 26/Jun/20


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