2log _x (3) log _(3x) (3)=log


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Question Number 161342 by bobhans last updated on 16/Dec/21

$2 \log_{x} (3) \log_{3 x} (3) = \log_{9 \sqrt{x}} (3)$ $x = ?$
	Answered by 1549442205PVT last updated on 16/Dec/21
	$the given equation is equivalent to (2/(log_3 x)).(1/(1+log_3 x))=(1/(2+(1/2)log_3 x))(for x>0,x≠1) ⇔2(4+log_3 x)=2.log_3 x(1+log_3 x) ⇔log_3 ^2 x=4⇔log_3 x=±2⇔x=3^(±2) ⇔x∈{(1/9),9}$
	$t h e g i v e n e q u a t i o n i s e q u i v a l e n t t o$ $\frac{2}{{l o g}_{3} x} . \frac{1}{1 + {l o g}_{3} x} = \frac{1}{2 + \frac{1}{2} {l o g}_{3} x} (f o r x > 0, x \neq 1)$ $\Leftrightarrow 2 (4 + {l o g}_{3} x) = 2 . {l o g}_{3} x (1 + {l o g}_{3} x)$ $\Leftrightarrow {l o g}_{3}^{2} x = 4 \Leftrightarrow {l o g}_{3} x = \pm 2 \Leftrightarrow x = 3^{\pm 2}$ $\Leftrightarrow x \in {\frac{1}{9}, 9}$
	Answered by cortano last updated on 16/Dec/21
	$(Q) 2 log _x (3) log _(3x) (3)= log _(9(√x)) (3) x=? (Sol) (2/(log _3 (x).(log _3 (x)+1))) = (1/((1/2)log _3 (x)+2)) (2/(log _3 (x)(log _3 (x)+1))) = (2/(log _3 (x)+4)) set log _3 (x)= y ⇔ y^2 +y = y+4 ⇔y=±2 → { ((log _3 (x)=2⇒x=9)),((log _3 (x)=−2⇒x=(1/9))) :}$
	$(Q) 2 \log_{x} (3) \log_{3 x} (3) = \log_{9 \sqrt{x}} (3)$ $x = ?$ $(S o l) \frac{2}{\log_{3} (x) . (\log_{3} (x) + 1)} = \frac{1}{\frac{1}{2} \log_{3} (x) + 2}$ $\frac{2}{\log_{3} (x) (\log_{3} (x) + 1)} = \frac{2}{\log_{3} (x) + 4}$ $s e t \log_{3} (x) = y$ $\Leftrightarrow y^{2} + y = y + 4$ $\Leftrightarrow y = \pm 2 \to {\begin{cases} \log_{3} (x) = 2 \Rightarrow x = 9 \\ \log_{3} (x) = - 2 \Rightarrow x = \frac{1}{9} \end{cases}$
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