(log_x (6))^2 + (log_(1/6) ((1/x)))^2 + log_(1/(√x)) ((1/6)) + log


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Question Number 89456 by jagoll last updated on 17/Apr/20

${(\log_{x} (6))}^{2} + {(\log_{\frac{1}{6}} (\frac{1}{x}))}^{2} +$ $\log_{\frac{1}{\sqrt{x}}} (\frac{1}{6}) + \log_{\sqrt{6}} (x) + \frac{3}{4} = 0$
	Answered by john santu last updated on 17/Apr/20
	$(1) (log_(1/6) ((1/x)))^2 = (log_6 (x))^2 (2) log_(1/(√x)) ((1/6)) =2 log_x (6) (3) log_(√6) (x) = 2 log_6 (x) ⇒let log_6 (x) = t t^2 + (1/t^2 ) + 2t + (2/t) + (3/4) = 0 (t+(1/t))^2 +2(t+(1/t))−(5/4) = 0 ⇒4u^2 + 8u −5 = 0 (2u−1)(2u+5) = 0 (i) 2t +(2/t)−1 = 0 2t^2 −t +2 = 0 , t ∉ R (ii) 2t + (2/t) +5 = 0 2t^2 + 5t +2 = 0 (2t +1) (t+2) = 0 { ((log_6 (x) = −2 ⇒ x = (1/(36)))),((log_6 (x) = −(1/2) ⇒x = (1/(√6)))) :}$
	$(1) {(\log_{\frac{1}{6}} (\frac{1}{x}))}^{2} = {(\log_{6} (x))}^{2}$ $(2) \log_{\frac{1}{\sqrt{x}}} (\frac{1}{6}) = 2 \log_{x} (6)$ $(3) \log_{\sqrt{6}} (x) = 2 \log_{6} (x)$ $\Rightarrow l e t \log_{6} (x) = t$ $t^{2} + \frac{1}{t^{2}} + 2 t + \frac{2}{t} + \frac{3}{4} = 0$ ${(t + \frac{1}{t})}^{2} + 2 (t + \frac{1}{t}) - \frac{5}{4} = 0$ $\Rightarrow 4 u^{2} + 8 u - 5 = 0$ $(2 u - 1) (2 u + 5) = 0$ $(i) 2 t + \frac{2}{t} - 1 = 0$ $2 t^{2} - t + 2 = 0, t \notin R$ $(i i) 2 t + \frac{2}{t} + 5 = 0$ $2 t^{2} + 5 t + 2 = 0$ $(2 t + 1) (t + 2) = 0$ ${\begin{cases} \log_{6} (x) = - 2 \Rightarrow x = \frac{1}{36} \\ \log_{6} (x) = - \frac{1}{2} \Rightarrow x = \frac{1}{\sqrt{6}} \end{cases}$
	Commented by jagoll last updated on 17/Apr/20

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