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Question Number 156109 by cortano last updated on 08/Oct/21

$$\log {}_5\left(\sqrt{\mathrm{x}-9}\right)-\log {}_5({3\mathrm{x}}^2-12)-\log {}_5\left(\sqrt{2\mathrm{x}-1}\right)⩽0$$

Answered by yeti123 last updated on 08/Oct/21

$$\left(1\right)\cap \left(2\right)\cap \left(3\right)\cap \left(4\right)\cap \left(5\right)\cap \left(6\right)=9<x⩽21478$$

Answered by yeti123 last updated on 08/Oct/21

$${\log }_5\left(\sqrt{x-9}\right)-{\log }_5(3x^2-12)-{\log }_5\left(\sqrt{2x-1}\right)⩽0$$$${\log }_5\left(\frac{\sqrt{x-9}}{(3x^2-12)\left(\sqrt{2x-1}\right)}\right)⩽{\log }_5\left(1\right)$$$$\frac{\sqrt{x-9}}{(3x^2-12)\sqrt{2x-1}}⩽1$$$$\Rightarrow x=9....\left(1\right)$$$$-2<x<\frac{1}{2}.....\left(2\right)$$$$x⩽21478........\left(3\right)$$$$$$$${\log }_5\left(\sqrt{x-9}\right)\Rightarrow \sqrt{x-9}>0\Rightarrow x>9..............\left(4\right)$$$${\log }_5(3x^2-12)\Rightarrow (3x^2-12)>0\Rightarrow x>2\wedge x<-2..........\left(5\right)$$$${\log }_5\left(\sqrt{2x-1}\right)\Rightarrow \sqrt{2x-1}>0\Rightarrow x>\frac{1}{2}............\left(6\right)$$$$$$$$\left(1\right)\cap \left(2\right)\cap \left(3\right)\cap \left(4\right)\cap \left(5\right)\cap \left(6\right)=\frac{1}{2}<x⩽21478$$

Commented by cortano last updated on 08/Oct/21

$$\mathrm{for}\mathrm{x}=1\mathrm{is}\mathrm{not}\mathrm{solution}$$

Commented by mr W last updated on 08/Oct/21

$$whyx⩽21478?howdidyougetit?$$$$infactwehavex>9andforanyx>9$$$$0<\frac{\sqrt{x-9}}{(3x^2-12)\sqrt{2x-1}}⩽0.0009⩽1$$$$$$$$sotheanswerisx\in (9,+\mathrm{\infty })$$

Commented by yeti123 last updated on 08/Oct/21

$$\mathrm{yes}\mathrm{mr}.\mathrm{W},\mathrm{I}\mathrm{miscalculated}.\mathrm{thanks}\mathrm{for}\mathrm{the}\mathrm{correction}.$$

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