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

2log _x (3) log _(3x) (3)=log _(9(√x)) (3) x=?

$$\:\:\mathrm{2log}\:_{\mathrm{x}} \left(\mathrm{3}\right)\:\mathrm{log}\:_{\mathrm{3x}} \left(\mathrm{3}\right)=\mathrm{log}\:_{\mathrm{9}\sqrt{\mathrm{x}}} \left(\mathrm{3}\right) \\ $$$$\:\mathrm{x}=? \\ $$

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}

$${the}\:{given}\:{equation}\:{is}\:{equivalent}\:{to} \\ $$$$\frac{\mathrm{2}}{{log}_{\mathrm{3}} {x}}.\frac{\mathrm{1}}{\mathrm{1}+{log}_{\mathrm{3}} {x}}=\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}{log}_{\mathrm{3}} {x}}\left({for}\:{x}>\mathrm{0},{x}\neq\mathrm{1}\right) \\ $$$$\Leftrightarrow\mathrm{2}\left(\mathrm{4}+{log}_{\mathrm{3}} {x}\right)=\mathrm{2}.{log}_{\mathrm{3}} {x}\left(\mathrm{1}+{log}_{\mathrm{3}} {x}\right) \\ $$$$\Leftrightarrow{log}_{\mathrm{3}} ^{\mathrm{2}} {x}=\mathrm{4}\Leftrightarrow{log}_{\mathrm{3}} {x}=\pm\mathrm{2}\Leftrightarrow{x}=\mathrm{3}^{\pm\mathrm{2}} \\ $$$$\Leftrightarrow{x}\in\left\{\frac{\mathrm{1}}{\mathrm{9}},\mathrm{9}\right\} \\ $$

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))) :}

$$\:\:\left({Q}\right)\:\mathrm{2}\:\mathrm{log}\:_{{x}} \left(\mathrm{3}\right)\:\mathrm{log}\:_{\mathrm{3}{x}} \left(\mathrm{3}\right)=\:\mathrm{log}\:_{\mathrm{9}\sqrt{{x}}} \left(\mathrm{3}\right) \\ $$$$\:\:\:\:{x}=? \\ $$$$\:\left({Sol}\right)\:\frac{\mathrm{2}}{\mathrm{log}\:_{\mathrm{3}} \left({x}\right).\left(\mathrm{log}\:_{\mathrm{3}} \left({x}\right)+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:_{\mathrm{3}} \left({x}\right)+\mathrm{2}} \\ $$$$\:\:\:\frac{\mathrm{2}}{\mathrm{log}\:_{\mathrm{3}} \left({x}\right)\left(\mathrm{log}\:_{\mathrm{3}} \left({x}\right)+\mathrm{1}\right)}\:=\:\frac{\mathrm{2}}{\mathrm{log}\:_{\mathrm{3}} \left({x}\right)+\mathrm{4}} \\ $$$$\:\:{set}\:\mathrm{log}\:_{\mathrm{3}} \left({x}\right)=\:{y} \\ $$$$\:\:\Leftrightarrow\:{y}^{\mathrm{2}} +{y}\:=\:{y}+\mathrm{4}\: \\ $$$$\:\:\Leftrightarrow{y}=\pm\mathrm{2}\:\rightarrow\begin{cases}{\mathrm{log}\:_{\mathrm{3}} \left({x}\right)=\mathrm{2}\Rightarrow{x}=\mathrm{9}}\\{\mathrm{log}\:_{\mathrm{3}} \left({x}\right)=−\mathrm{2}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{9}}}\end{cases} \\ $$

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