Menu Close

log-5-x-9-log-5-3x-2-12-log-5-2x-1-0-




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

Leave a Reply

Your email address will not be published. Required fields are marked *