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Question Number 32999 by artibunja last updated on 09/Apr/18

Commented by prof Abdo imad last updated on 09/Apr/18

$\Leftrightarrow {(\frac{l n (x + 1)}{l n (\frac{1}{3})})}^{- 1} > 1 \Leftrightarrow {(\frac{l n (x + 1)}{- l n (3)})}^{- 1} > 1 w i t h x > - 1$ $\Leftrightarrow - \frac{l n 3}{l n (x + 1)} > 1 \Leftrightarrow \frac{l n 3}{l n (x + 1)} < - 1$ $\Leftrightarrow \frac{l n 3}{l n (x + 1)} + 1 < 0 \Leftrightarrow \frac{l n (x + 1) + l n (3)}{l n (x + 1)} < 0$ $c a s e 1 x > 0 \Rightarrow x + 1 > 1 \Rightarrow l n (x + 1) > 0 (e) \Leftrightarrow$ $l n (3 x + 3) < 0 \Leftrightarrow 0 < 3 x + 3 < 1 \Rightarrow - 1 < x < \frac{- 2}{3}$ $b u t x > 0 s o n o s o l u t i o n$ $c a s e 2 - 1 < x < 0 (e) \Leftrightarrow l n (3 x + 3) > 0 \Leftrightarrow 3 x + 3 > 1$ $\Leftrightarrow x > - \frac{2}{3} \Rightarrow S_{2} =] - \frac{2}{3}, 0 [\Rightarrow S = S_{1} \cup S_{2}$ $=] - \frac{2}{3}, 0 [.$
	Answered by artibunja last updated on 09/Apr/18

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