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

$$\frac{1}{2}+{log}_{9}x-{log}_{3}5x>{log}_{\frac{1}{3}}(x+1)$$

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

$$\frac{1}{2}+{log}_{9}x-{log}_{3}5x>{log}_{\frac{1}{3}}(x+1)$$ $$\frac{1}{2}+\frac{{log}_{3}x}{{log}_{3}9}-\frac{{log}_{3}5x}{{log}_{3}3}>\frac{{log}_3(x+1)}{{log}_3\frac{1}{3}}$$ $$\frac{1}{2}+\frac{{log}_{3}x}{2}-\frac{{log}_{3}5x}{1}>\frac{{log}_3(x+1)}{-1}$$ $$1+{log}_{3}x-2{log}_{3}5x>-2{log}_3(x+1)$$ $$\underset{-}{{log}_{3}3+{log}_{3}x}-{log}_3{\left(5x\right)}^2>{log}_3{(x+1)}^{-2}$$ $$\underset{-}{{log}_{3}3x}-{log}_{3}25x^2>{log}_3{(x+1)}^{-2}$$ $${log}_3\left(\frac{3x}{25x^2}\right)>{log}_3\left(\frac{1}{{(x+1)}^2}\right)$$ $$\frac{3}{25x}>\frac{1}{{(x+1)}^2}$$ $$\frac{3}{25x}-\frac{1}{{(x+1)}^2}>0$$ $$\frac{3{(x+1)}^2-25x}{25x{(1+x)}^2}>0$$ $$\frac{3x^2+6x+3-25x}{25x{(1+x)}^2}>0$$ $$\frac{3x^2-19x+3}{25x{(1+x)}^2}>0$$ $$\frac{N}{D}>0\Rightarrow N,D>0\mid N,D<0$$ $$WhenN,D>0$$ $$3x^2-19x+3>0\wedge 25x{(1+x)}^2>0$$ $$25x{(1+x)}^2>0\Rightarrow x>0$$ $$$$ $$$$ $$WhenN,D<0$$ $$3x^2-19x+3<0\wedge 25x{(1+x)}^2<0$$ $$$$ $$Continue$$

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