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

log _5 ((√(x−9)))−log _5 (3x^2 −12)−log _5 ((√(2x−1))) ≤ 0

$$\:\:\mathrm{log}\:_{\mathrm{5}} \left(\sqrt{\mathrm{x}−\mathrm{9}}\right)−\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{3x}^{\mathrm{2}} −\mathrm{12}\right)−\mathrm{log}\:_{\mathrm{5}} \left(\sqrt{\mathrm{2x}−\mathrm{1}}\right)\:\leqslant\:\mathrm{0} \\ $$

Answered by yeti123 last updated on 08/Oct/21

(1) ∩ (2) ∩ (3) ∩ (4) ∩ (5) ∩ (6) = 9 < x ≤ 21478

$$\left(\mathrm{1}\right)\:\cap\:\left(\mathrm{2}\right)\:\cap\:\left(\mathrm{3}\right)\:\cap\:\left(\mathrm{4}\right)\:\cap\:\left(\mathrm{5}\right)\:\cap\:\left(\mathrm{6}\right)\:=\:\mathrm{9}\:<\:{x}\:\leqslant\:\mathrm{21478} \\ $$

Answered by yeti123 last updated on 08/Oct/21

log_5 ((√(x−9))) − log_5 (3x^2 −12) − log_5 ((√(2x−1))) ≤ 0 log_5 (((√(x−9))/((3x^2 −12)((√(2x−1)))))) ≤ log_5 (1) ((√(x−9))/((3x^2 −12)(√(2x−1)))) ≤ 1 ⇒ x = 9 ....(1) −2 < x <(1/2).....(2) x ≤ 21478........(3) log_5 ((√(x−9))) ⇒ (√(x−9 )) > 0 ⇒ x> 9..............(4) log_5 (3x^2 −12) ⇒ (3x^2 −12) > 0 ⇒ x > 2 ∧ x < −2..........(5) log_5 ((√(2x−1))) ⇒ (√(2x−1)) > 0 ⇒ x > (1/2)............(6) (1) ∩ (2) ∩ (3) ∩ (4) ∩ (5) ∩ (6) = (1/2) < x ≤ 21478

$$\mathrm{log}_{\mathrm{5}} \left(\sqrt{{x}−\mathrm{9}}\right)\:−\:\mathrm{log}_{\mathrm{5}} \left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}\right)\:−\:\mathrm{log}_{\mathrm{5}} \left(\sqrt{\mathrm{2}{x}−\mathrm{1}}\right)\:\leqslant\:\mathrm{0} \\ $$$$\mathrm{log}_{\mathrm{5}} \left(\frac{\sqrt{{x}−\mathrm{9}}}{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}\right)\left(\sqrt{\mathrm{2}{x}−\mathrm{1}}\right)}\right)\:\leqslant\:\mathrm{log}_{\mathrm{5}} \left(\mathrm{1}\right) \\ $$$$\frac{\sqrt{{x}−\mathrm{9}}}{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}\right)\sqrt{\mathrm{2}{x}−\mathrm{1}}}\:\leqslant\:\mathrm{1} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{9}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:−\mathrm{2}\:<\:{x}\:<\frac{\mathrm{1}}{\mathrm{2}}.....\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:{x}\:\leqslant\:\mathrm{21478}........\left(\mathrm{3}\right) \\ $$$$ \\ $$$$\mathrm{log}_{\mathrm{5}} \left(\sqrt{{x}−\mathrm{9}}\right)\:\Rightarrow\:\sqrt{{x}−\mathrm{9}\:}\:>\:\mathrm{0}\:\Rightarrow\:{x}>\:\mathrm{9}..............\left(\mathrm{4}\right) \\ $$$$\mathrm{log}_{\mathrm{5}} \left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}\right)\:\Rightarrow\:\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}\right)\:>\:\mathrm{0}\:\Rightarrow\:{x}\:>\:\mathrm{2}\:\wedge\:{x}\:<\:−\mathrm{2}..........\left(\mathrm{5}\right) \\ $$$$\mathrm{log}_{\mathrm{5}} \left(\sqrt{\mathrm{2}{x}−\mathrm{1}}\right)\:\Rightarrow\:\sqrt{\mathrm{2}{x}−\mathrm{1}}\:>\:\mathrm{0}\:\Rightarrow\:{x}\:>\:\frac{\mathrm{1}}{\mathrm{2}}............\left(\mathrm{6}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\cap\:\left(\mathrm{2}\right)\:\cap\:\left(\mathrm{3}\right)\:\cap\:\left(\mathrm{4}\right)\:\cap\:\left(\mathrm{5}\right)\:\cap\:\left(\mathrm{6}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:<\:{x}\:\leqslant\:\mathrm{21478} \\ $$

Commented by cortano last updated on 08/Oct/21

for x=1 is not solution

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

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

why x≤21478? how did you get it? in fact we have x>9 and for any x>9 0<((√(x−9))/((3x^2 −12)(√(2x−1)))) ≤0.0009 ≤1 so the answer is x∈(9,+∞)

$${why}\:{x}\leqslant\mathrm{21478}?\:{how}\:{did}\:{you}\:{get}\:{it}? \\ $$$${in}\:{fact}\:{we}\:{have}\:{x}>\mathrm{9}\:{and}\:{for}\:{any}\:{x}>\mathrm{9} \\ $$$$\mathrm{0}<\frac{\sqrt{{x}−\mathrm{9}}}{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}\right)\sqrt{\mathrm{2}{x}−\mathrm{1}}}\:\leqslant\mathrm{0}.\mathrm{0009}\:\leqslant\mathrm{1} \\ $$$$ \\ $$$${so}\:{the}\:{answer}\:{is}\:{x}\in\left(\mathrm{9},+\infty\right) \\ $$

Commented by yeti123 last updated on 08/Oct/21

yes mr. W, I miscalculated. thanks for the correction.

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

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