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Question Number 161786 by mathlove last updated on 22/Dec/21

logx_(4x) +logx_(x/2) =2 solve for x=?

$${log}\underset{\mathrm{4}{x}} {{x}}+{log}\underset{\frac{{x}}{\mathrm{2}}} {{x}}=\mathrm{2} \\ $$$${solve}\:\:\:{for}\:\:\:{x}=? \\ $$

Answered by mr W last updated on 22/Dec/21

((ln x)/(ln (4x)))+((ln x)/(ln ((x/2))))=2 ((ln x)/(2ln 2+ln x))+((ln x)/(ln x−ln 2))=2 let t=ln x, a=ln 2 (t/(2a+t))+(t/(t−a))=2 t=4a ln x=4 ln 2 ⇒x=2^4 =16

$$\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\left(\mathrm{4}{x}\right)}+\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\left(\frac{{x}}{\mathrm{2}}\right)}=\mathrm{2} \\ $$$$\frac{\mathrm{ln}\:{x}}{\mathrm{2ln}\:\mathrm{2}+\mathrm{ln}\:{x}}+\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:{x}−\mathrm{ln}\:\mathrm{2}}=\mathrm{2} \\ $$$${let}\:{t}=\mathrm{ln}\:{x},\:{a}=\mathrm{ln}\:\mathrm{2} \\ $$$$\frac{{t}}{\mathrm{2}{a}+{t}}+\frac{{t}}{{t}−{a}}=\mathrm{2} \\ $$$${t}=\mathrm{4}{a} \\ $$$$\mathrm{ln}\:{x}=\mathrm{4}\:\mathrm{ln}\:\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{2}^{\mathrm{4}} =\mathrm{16} \\ $$

Answered by cortano last updated on 22/Dec/21

(Q) Solve for x : log _(4x) (x)+log _(x/2) (x) = 2 (S) ⇒ ((log _2 (x))/(2+log _2 (x))) + ((log _2 (x))/(log _2 (x)−1)) = 2 [ z = log _2 (x); x>0 ; x≠1 ] ⇒ (z/(z+2)) + (z/(z−1)) = 2 ⇒z(z−1+z+2)= 2(z+2)(z−1) ⇒2z^2 +z = 2z^2 +2z−4 ⇒z = 4= log _2 (x) ⇒x = 2^4 = 16

$$\:\left({Q}\right)\:{Solve}\:{for}\:{x}\::\:\mathrm{log}\:_{\mathrm{4}{x}} \left({x}\right)+\mathrm{log}\:_{\frac{{x}}{\mathrm{2}}} \left({x}\right)\:=\:\mathrm{2} \\ $$$$\:\left({S}\right)\:\Rightarrow\:\frac{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)}{\mathrm{2}+\mathrm{log}\:_{\mathrm{2}} \left({x}\right)}\:+\:\frac{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)}{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)−\mathrm{1}}\:=\:\mathrm{2} \\ $$$$\:\left[\:{z}\:=\:\mathrm{log}\:_{\mathrm{2}} \left({x}\right);\:{x}>\mathrm{0}\:;\:{x}\neq\mathrm{1}\:\right] \\ $$$$\:\Rightarrow\:\frac{{z}}{{z}+\mathrm{2}}\:+\:\frac{{z}}{{z}−\mathrm{1}}\:=\:\mathrm{2} \\ $$$$\Rightarrow{z}\left({z}−\mathrm{1}+{z}+\mathrm{2}\right)=\:\mathrm{2}\left({z}+\mathrm{2}\right)\left({z}−\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{2}{z}^{\mathrm{2}} +{z}\:=\:\mathrm{2}{z}^{\mathrm{2}} \:+\mathrm{2}{z}−\mathrm{4} \\ $$$$\:\Rightarrow{z}\:=\:\mathrm{4}=\:\mathrm{log}\:_{\mathrm{2}} \left({x}\right) \\ $$$$\Rightarrow{x}\:=\:\mathrm{2}^{\mathrm{4}} =\:\mathrm{16} \\ $$

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