log-2x-1-x-1-gt-log-4-2x-x-1-x-

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 by Nimatullah last updated on 25/Feb/22

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Answered by mahdipoor last updated on 25/Feb/22

\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}