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 4592 by love math last updated on 09/Feb/16

(1/2)+log_9 x−log_3 5x>log_(1/3) (x+1)

$$\frac{\mathrm{1}}{\mathrm{2}}+{log}_{\mathrm{9}} {x}−{log}_{\mathrm{3}} \mathrm{5}{x}>{log}_{\frac{\mathrm{1}}{\mathrm{3}}} \left({x}+\mathrm{1}\right) \\ $$

Answered by Rasheed Soomro last updated on 10/Feb/16

(1/2)+log_9 x−log_3 5x>log_(1/3) (x+1) (1/2)+((log_3 x)/(log_3 9))−((log_3 5x)/(log_3 3))>((log_3 (x+1))/(log_3 (1/3))) (1/2)+((log_3 x)/2)−((log_3 5x)/1)>((log_3 (x+1))/(−1)) 1+log_3 x−2log_3 5x>−2log_3 (x+1) log_3 3+log_3 x_(−) −log_3 (5x)^2 >log_3 (x+1)^(−2) log_3 3x_(−) −log_3 25x^2 >log_3 (x+1)^(−2) log_3 (((3x)/(25x^2 )))>log_3 ((1/((x+1)^2 ))) (3/(25x))>(1/((x+1)^2 )) (3/(25x))−(1/((x+1)^2 ))>0 ((3(x+1)^2 −25x)/(25x(1+x)^2 ))>0 ((3x^2 +6x+3−25x)/(25x(1+x)^2 ))>0 ((3x^2 −19x+3)/(25x(1+x)^2 ))>0 (N/D)>0⇒ N ,D >0 ∣ N ,D<0 When N,D>0 3x^2 −19x+3>0 ∧ 25x(1+x)^2 >0 25x(1+x)^2 >0⇒x>0 When N,D<0 3x^2 −19x+3<0 ∧ 25x(1+x)^2 <0 Continue

$$\frac{\mathrm{1}}{\mathrm{2}}+{log}_{\mathrm{9}} {x}−{log}_{\mathrm{3}} \mathrm{5}{x}>{log}_{\frac{\mathrm{1}}{\mathrm{3}}} \left({x}+\mathrm{1}\right) \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}}+\frac{{log}_{\mathrm{3}} \:{x}}{{log}_{\mathrm{3}} \:\mathrm{9}}−\frac{{log}_{\mathrm{3}} \:\mathrm{5}{x}}{{log}_{\mathrm{3}} \:\mathrm{3}}>\frac{{log}_{\mathrm{3}} \:\left({x}+\mathrm{1}\right)}{{log}_{\mathrm{3}} \:\frac{\mathrm{1}}{\mathrm{3}}} \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}}+\frac{{log}_{\mathrm{3}} \:{x}}{\mathrm{2}}−\frac{{log}_{\mathrm{3}} \:\mathrm{5}{x}}{\mathrm{1}}>\frac{{log}_{\mathrm{3}} \:\left({x}+\mathrm{1}\right)}{−\mathrm{1}} \\ $$ $$\mathrm{1}+{log}_{\mathrm{3}} {x}−\mathrm{2}{log}_{\mathrm{3}} \mathrm{5}{x}>−\mathrm{2}{log}_{\mathrm{3}} \left({x}+\mathrm{1}\right) \\ $$ $$\underset{−} {{log}_{\mathrm{3}} \mathrm{3}+{log}_{\mathrm{3}} {x}}−{log}_{\mathrm{3}} \left(\mathrm{5}{x}\right)^{\mathrm{2}} >{log}_{\mathrm{3}} \left({x}+\mathrm{1}\right)^{−\mathrm{2}} \\ $$ $$\underset{−} {{log}_{\mathrm{3}} \mathrm{3}{x}}−{log}_{\mathrm{3}} \mathrm{25}{x}^{\mathrm{2}} >{log}_{\mathrm{3}} \left({x}+\mathrm{1}\right)^{−\mathrm{2}} \\ $$ $${log}_{\mathrm{3}} \left(\frac{\mathrm{3}{x}}{\mathrm{25}{x}^{\mathrm{2}} }\right)>{log}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$ $$\frac{\mathrm{3}}{\mathrm{25}{x}}>\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ $$\frac{\mathrm{3}}{\mathrm{25}{x}}−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }>\mathrm{0} \\ $$ $$\frac{\mathrm{3}\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{25}{x}}{\mathrm{25}{x}\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }>\mathrm{0} \\ $$ $$\frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{3}−\mathrm{25}{x}}{\mathrm{25}{x}\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }>\mathrm{0} \\ $$ $$\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{19}{x}+\mathrm{3}}{\mathrm{25}{x}\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }>\mathrm{0} \\ $$ $$\frac{{N}}{{D}}>\mathrm{0}\Rightarrow\:{N}\:,{D}\:>\mathrm{0}\:\mid\:{N}\:,{D}<\mathrm{0}\: \\ $$ $${When}\:{N},{D}>\mathrm{0} \\ $$ $$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{19}{x}+\mathrm{3}>\mathrm{0}\:\wedge\:\mathrm{25}{x}\left(\mathrm{1}+{x}\right)^{\mathrm{2}} >\mathrm{0} \\ $$ $$\mathrm{25}{x}\left(\mathrm{1}+{x}\right)^{\mathrm{2}} >\mathrm{0}\Rightarrow{x}>\mathrm{0} \\ $$ $$ \\ $$ $$ \\ $$ $${When}\:{N},{D}<\mathrm{0} \\ $$ $$\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{19}{x}+\mathrm{3}<\mathrm{0}\:\:\wedge\:\mathrm{25}{x}\left(\mathrm{1}+{x}\right)^{\mathrm{2}} <\mathrm{0} \\ $$ $$ \\ $$ $${Continue} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com