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Question Number 166687 by cortano1 last updated on 25/Feb/22
log _((2x−1)) (x+1) > log _((4−2x)) (x+1) x=?
$$\:\:\:\mathrm{log}\:_{\left(\mathrm{2x}−\mathrm{1}\right)} \left(\mathrm{x}+\mathrm{1}\right)\:>\:\mathrm{log}\:_{\left(\mathrm{4}−\mathrm{2x}\right)} \left(\mathrm{x}+\mathrm{1}\right) \\ $$$$\:\:\mathrm{x}=? \\ $$
Commented by Nimatullah last updated on 25/Feb/22
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$${p} \\ $$
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}
$$\Rightarrow{D}_{{eq}} =\:\mathrm{0}.\mathrm{5}<{x}<\mathrm{2}\:\:{x}\neq\mathrm{1},\mathrm{1}.\mathrm{5} \\ $$$$\frac{{ln}\left({x}+\mathrm{1}\right)}{{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)}>\frac{{ln}\left({x}+\mathrm{1}\right)}{{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)}\:\Rightarrow{ln}\left({x}+\mathrm{1}\right)>\mathrm{0}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)}>\frac{\mathrm{1}}{{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)}\:\:\:\left({A}\right) \\ $$$$\left.{I}\right)\mathrm{0}.\mathrm{5}<{x}<\mathrm{1}\:\Rightarrow{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)<\mathrm{0}\:,\:{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)>\mathrm{0} \\ $$$$\Rightarrow{A}\Rightarrow\:\mathrm{0}>\frac{\mathrm{1}}{{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)}>\frac{\mathrm{1}}{{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)}>\mathrm{0}\Rightarrow\nexists{x} \\ $$$$\left.{II}\right)\mathrm{1}<{x}<\mathrm{1}.\mathrm{5}\:\Rightarrow{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)>\mathrm{0}\:,\:{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)>\mathrm{0} \\ $$$$\Rightarrow{A}\Rightarrow{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)<{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)\Rightarrow\mathrm{2}{x}−\mathrm{1}<\mathrm{4}−\mathrm{2}{x} \\ $$$$\Rightarrow{x}<\mathrm{2}.\mathrm{5}\:\overset{\mathrm{1}<{x}<\mathrm{1}.\mathrm{5}} {\Rightarrow}\:{x}\in\left(\mathrm{1},\mathrm{1}.\mathrm{5}\right) \\ $$$$\left.{III}\right)\mathrm{1}.\mathrm{5}<{x}<\mathrm{2}\Rightarrow{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)>\mathrm{0}\:,\:{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)<\mathrm{0} \\ $$$$\Rightarrow{A}\Rightarrow\:\frac{\mathrm{1}}{{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)}>\mathrm{0}>\:\frac{\mathrm{1}}{{ln}\left(\mathrm{4}{x}−\mathrm{1}\right)} \\ $$$$\Rightarrow{x}\in{D}_{{eq}} \overset{\mathrm{1}.\mathrm{5}<{x}<\mathrm{2}} {\Rightarrow}{x}\in\left(\mathrm{1}.\mathrm{5},\mathrm{2}\right) \\ $$$${Ans}={I}\cup{II}\cup{III}=\left(\mathrm{1},\mathrm{2}\right)−\left\{\mathrm{1}.\mathrm{5}\right\} \\ $$$$ \\ $$

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