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Question Number 89456 by jagoll last updated on 17/Apr/20
(log_x (6))^2 + (log_(1/6) ((1/x)))^2 + log_(1/( (√x))) ((1/6)) + log_(√6) (x) + (3/4) = 0
$$\left(\mathrm{log}_{{x}} \left(\mathrm{6}\right)\right)^{\mathrm{2}} \:+\:\left(\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{6}}} \left(\frac{\mathrm{1}}{{x}}\right)\right)^{\mathrm{2}} +\: \\ $$$$\mathrm{log}_{\frac{\mathrm{1}}{\:\sqrt{{x}}}} \left(\frac{\mathrm{1}}{\mathrm{6}}\right)\:+\:\mathrm{log}_{\sqrt{\mathrm{6}}} \:\left({x}\right)\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:=\:\mathrm{0} \\ $$
Answered by john santu last updated on 17/Apr/20
(1) (log_(1/6) ((1/x)))^2 = (log_6 (x))^2 (2) log_(1/( (√x))) ((1/6)) =2 log_x (6) (3) log_(√6) (x) = 2 log_6 (x) ⇒let log_6 (x) = t t^2 + (1/t^2 ) + 2t + (2/t) + (3/4) = 0 (t+(1/t))^2 +2(t+(1/t))−(5/4) = 0 ⇒4u^2 + 8u −5 = 0 (2u−1)(2u+5) = 0 (i) 2t +(2/t)−1 = 0 2t^2 −t +2 = 0 , t ∉ R (ii) 2t + (2/t) +5 = 0 2t^2 + 5t +2 = 0 (2t +1) (t+2) = 0 { ((log_6 (x) = −2 ⇒ x = (1/(36)))),((log_6 (x) = −(1/2) ⇒x = (1/( (√6))))) :}
$$\left(\mathrm{1}\right)\:\left(\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{6}}} \left(\frac{\mathrm{1}}{{x}}\right)\right)^{\mathrm{2}} \:=\:\left(\mathrm{log}_{\mathrm{6}} \left({x}\right)\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{log}_{\frac{\mathrm{1}}{\:\sqrt{{x}}}} \:\left(\frac{\mathrm{1}}{\mathrm{6}}\right)\:=\mathrm{2}\:\mathrm{log}_{{x}} \left(\mathrm{6}\right) \\ $$$$\left(\mathrm{3}\right)\:\mathrm{log}_{\sqrt{\mathrm{6}}} \:\left({x}\right)\:=\:\mathrm{2}\:\mathrm{log}_{\mathrm{6}} \left({x}\right) \\ $$$$\Rightarrow{let}\:\mathrm{log}_{\mathrm{6}} \left({x}\right)\:=\:{t} \\ $$$${t}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:+\:\mathrm{2}{t}\:+\:\frac{\mathrm{2}}{{t}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:=\:\mathrm{0} \\ $$$$\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}\left({t}+\frac{\mathrm{1}}{{t}}\right)−\frac{\mathrm{5}}{\mathrm{4}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}{u}^{\mathrm{2}} \:+\:\mathrm{8}{u}\:−\mathrm{5}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}{u}−\mathrm{1}\right)\left(\mathrm{2}{u}+\mathrm{5}\right)\:=\:\mathrm{0} \\ $$$$\left({i}\right)\:\mathrm{2}{t}\:+\frac{\mathrm{2}}{{t}}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} \:−{t}\:+\mathrm{2}\:=\:\mathrm{0}\:,\:{t}\:\notin\:\mathbb{R} \\ $$$$\left({ii}\right)\:\mathrm{2}{t}\:+\:\frac{\mathrm{2}}{{t}}\:+\mathrm{5}\:=\:\mathrm{0} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} \:+\:\mathrm{5}{t}\:+\mathrm{2}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}{t}\:+\mathrm{1}\right)\:\left({t}+\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$$\begin{cases}{\mathrm{log}_{\mathrm{6}} \left({x}\right)\:=\:−\mathrm{2}\:\Rightarrow\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{36}}}\\{\mathrm{log}_{\mathrm{6}} \left({x}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{x}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}}\end{cases} \\ $$
Commented by jagoll last updated on 17/Apr/20
great sir. thank you
$${great}\:{sir}.\:{thank}\:{you} \\ $$

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