Question and Answers Forum

All Questions      Topic List

Logarithms Questions

Previous in All Question      Next in All Question      

Previous in Logarithms      Next in Logarithms      

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 byNimatullah last updated on 25/Feb/22

p

$${p} \\ $$

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}

$$\Rightarrow{D}_{{eq}} =\:\mathrm{0}.\mathrm{5}<{x}<\mathrm{2}\:\:{x}\neq\mathrm{1},\mathrm{1}.\mathrm{5} \\ $$ $$\frac{{ln}\left({x}+\mathrm{1}\right)}{{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)}>\frac{{ln}\left({x}+\mathrm{1}\right)}{{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)}\:\Rightarrow{ln}\left({x}+\mathrm{1}\right)>\mathrm{0}\:\Rightarrow \\ $$ $$\frac{\mathrm{1}}{{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)}>\frac{\mathrm{1}}{{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)}\:\:\:\left({A}\right) \\ $$ $$\left.{I}\right)\mathrm{0}.\mathrm{5}<{x}<\mathrm{1}\:\Rightarrow{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)<\mathrm{0}\:,\:{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)>\mathrm{0} \\ $$ $$\Rightarrow{A}\Rightarrow\:\mathrm{0}>\frac{\mathrm{1}}{{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)}>\frac{\mathrm{1}}{{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)}>\mathrm{0}\Rightarrow\nexists{x} \\ $$ $$\left.{II}\right)\mathrm{1}<{x}<\mathrm{1}.\mathrm{5}\:\Rightarrow{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)>\mathrm{0}\:,\:{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)>\mathrm{0} \\ $$ $$\Rightarrow{A}\Rightarrow{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)<{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)\Rightarrow\mathrm{2}{x}−\mathrm{1}<\mathrm{4}−\mathrm{2}{x} \\ $$ $$\Rightarrow{x}<\mathrm{2}.\mathrm{5}\:\overset{\mathrm{1}<{x}<\mathrm{1}.\mathrm{5}} {\Rightarrow}\:{x}\in\left(\mathrm{1},\mathrm{1}.\mathrm{5}\right) \\ $$ $$\left.{III}\right)\mathrm{1}.\mathrm{5}<{x}<\mathrm{2}\Rightarrow{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)>\mathrm{0}\:,\:{ln}\left(\mathrm{4}−\mathrm{2}{x}\right)<\mathrm{0} \\ $$ $$\Rightarrow{A}\Rightarrow\:\frac{\mathrm{1}}{{ln}\left(\mathrm{2}{x}−\mathrm{1}\right)}>\mathrm{0}>\:\frac{\mathrm{1}}{{ln}\left(\mathrm{4}{x}−\mathrm{1}\right)} \\ $$ $$\Rightarrow{x}\in{D}_{{eq}} \overset{\mathrm{1}.\mathrm{5}<{x}<\mathrm{2}} {\Rightarrow}{x}\in\left(\mathrm{1}.\mathrm{5},\mathrm{2}\right) \\ $$ $${Ans}={I}\cup{II}\cup{III}=\left(\mathrm{1},\mathrm{2}\right)−\left\{\mathrm{1}.\mathrm{5}\right\} \\ $$ $$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com