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1-2-log-9-x-log-3-5x-gt-log-1-3-x-1-

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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)
12+log9x−log35x>log13(x+1)
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
12+log9x−log35x>log13(x+1)12+log3xlog39−log35xlog33>log3(x+1)log31312+log3x2−log35x1>log3(x+1)−11+log3x−2log35x>−2log3(x+1)log33+log3x−−log3(5x)2>log3(x+1)−2log33x−−log325x2>log3(x+1)−2log3(3x25x2)>log3(1(x+1)2)325x>1(x+1)2325x−1(x+1)2>03(x+1)2−25x25x(1+x)2>03x2+6x+3−25x25x(1+x)2>03x2−19x+325x(1+x)2>0ND>0⇒N,D>0∣N,D<0WhenN,D>03x2−19x+3>0∧25x(1+x)2>025x(1+x)2>0⇒x>0WhenN,D<03x2−19x+3<0∧25x(1+x)2<0Continue

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