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Question Number 166687 by cortano1 last updated on 25/Feb/22

$\log_{(2 x - 1)} (x + 1) > \log_{(4 - 2 x)} (x + 1)$ $x = ?$
Commented byNimatullah last updated on 25/Feb/22

$p$
	Answered by mahdipoor last updated on 25/Feb/22
	$⇒D_(eq) = 0.5<x<2 x≠1,1.5 ((ln(x+1))/(ln(2x−1)))>((ln(x+1))/(ln(4−2x))) ⇒ln(x+1)>0 ⇒ (1/(ln(2x−1)))>(1/(ln(4−2x))) (A) I)0.5<x<1 ⇒ln(2x−1)<0 , ln(4−2x)>0 ⇒A⇒ 0>(1/(ln(2x−1)))>(1/(ln(4−2x)))>0⇒∄x II)1<x<1.5 ⇒ln(2x−1)>0 , ln(4−2x)>0 ⇒A⇒ln(2x−1)<ln(4−2x)⇒2x−1<4−2x ⇒x<2.5 ⇒^(1<x<1.5) x∈(1,1.5) III)1.5<x<2⇒ln(2x−1)>0 , ln(4−2x)<0 ⇒A⇒ (1/(ln(2x−1)))>0> (1/(ln(4x−1))) ⇒x∈D_(eq) ⇒^(1.5<x<2) x∈(1.5,2) Ans=I∪II∪III=(1,2)−{1.5}$
	$\Rightarrow D_{e q} = 0.5 < x < 2 x \neq 1, 1.5$ $\frac{l n (x + 1)}{l n (2 x - 1)} > \frac{l n (x + 1)}{l n (4 - 2 x)} \Rightarrow l n (x + 1) > 0 \Rightarrow$ $\frac{1}{l n (2 x - 1)} > \frac{1}{l n (4 - 2 x)} (A)$ $I) 0.5 < x < 1 \Rightarrow l n (2 x - 1) < 0, l n (4 - 2 x) > 0$ $\Rightarrow A \Rightarrow 0 > \frac{1}{l n (2 x - 1)} > \frac{1}{l n (4 - 2 x)} > 0 \Rightarrow ∄ x$ $I I) 1 < x < 1.5 \Rightarrow l n (2 x - 1) > 0, l n (4 - 2 x) > 0$ $\Rightarrow A \Rightarrow l n (2 x - 1) < l n (4 - 2 x) \Rightarrow 2 x - 1 < 4 - 2 x$ $\Rightarrow x < 2.5 \overset{1 < x < 1.5}{\Rightarrow} x \in (1, 1.5)$ $I I I) 1.5 < x < 2 \Rightarrow l n (2 x - 1) > 0, l n (4 - 2 x) < 0$ $\Rightarrow A \Rightarrow \frac{1}{l n (2 x - 1)} > 0 > \frac{1}{l n (4 x - 1)}$ $\Rightarrow x \in D_{e q} \overset{1.5 < x < 2}{\Rightarrow} x \in (1.5, 2)$ $A n s = I \cup I I \cup I I I = (1, 2) - {1.5}$
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