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Question Number 222104 by efronzo1 last updated on 17/Jun/25
log _4 x − log _x^2 8 = 1 x =?
$$\:\:\:\:\mathrm{log}\:_{\mathrm{4}} \:\mathrm{x}\:−\:\mathrm{log}\:_{\mathrm{x}^{\mathrm{2}} } \:\mathrm{8}\:=\:\mathrm{1} \\ $$$$\:\:\:\:\mathrm{x}\:=?\: \\ $$
Answered by mr W last updated on 17/Jun/25
((ln x)/(ln 2^2 ))−((ln 2^3 )/(ln x^2 ))=1 ((ln x)/(ln 2))−((3ln 2)/(ln x))=2 t−(3/t)=2 with t=((ln x)/(ln 2)) t^2 −2t−3=0 (t+1)(t−3)=0 ⇒((ln x)/(ln 2))=t=−1 ∨ 3 ⇒ln x=−ln 2=ln (1/2) ∨ 3 ln 2=ln 8 ⇒x=(1/2) ∨ 8
$$\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{2}^{\mathrm{2}} }−\frac{\mathrm{ln}\:\mathrm{2}^{\mathrm{3}} }{\mathrm{ln}\:{x}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{2}}−\frac{\mathrm{3ln}\:\mathrm{2}}{\mathrm{ln}\:{x}}=\mathrm{2}\: \\ $$$${t}−\frac{\mathrm{3}}{{t}}=\mathrm{2}\:{with}\:{t}=\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{2}} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{3}=\mathrm{0} \\ $$$$\left({t}+\mathrm{1}\right)\left({t}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{2}}={t}=−\mathrm{1}\:\vee\:\mathrm{3} \\ $$$$\Rightarrow\mathrm{ln}\:{x}=−\mathrm{ln}\:\mathrm{2}=\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}}\:\vee\:\mathrm{3}\:\mathrm{ln}\:\mathrm{2}=\mathrm{ln}\:\mathrm{8} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\:\vee\:\mathrm{8} \\ $$
Answered by gregori last updated on 19/Jun/25
(1/2) log _2 (x)−(3/2)log _x (2)= 1 log _2 (x)−3.log _x (2)= 2 (log _2 (x))^2 −2.log _2 (x)−3=0 (log _2 (x)−3)(log _2 (x)+1)=0 { ((log _2 (x)=3⇒x=2^3 =8)),((log _2 (x)=−1⇒x=2^(−1) =(1/2))) :}
$$\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{log}\:_{\mathrm{2}} \left({x}\right)−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{log}\:_{{x}} \left(\mathrm{2}\right)=\:\mathrm{1} \\ $$$$\:\:\:\mathrm{log}\:_{\mathrm{2}} \left({x}\right)−\mathrm{3}.\mathrm{log}\:_{{x}} \left(\mathrm{2}\right)=\:\mathrm{2} \\ $$$$\:\:\:\left(\mathrm{log}\:_{\mathrm{2}} \left({x}\right)\right)^{\mathrm{2}} −\mathrm{2}.\mathrm{log}\:_{\mathrm{2}} \left({x}\right)−\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\left(\mathrm{log}\:_{\mathrm{2}} \left({x}\right)−\mathrm{3}\right)\left(\mathrm{log}\:_{\mathrm{2}} \left({x}\right)+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\begin{cases}{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)=\mathrm{3}\Rightarrow{x}=\mathrm{2}^{\mathrm{3}} =\mathrm{8}}\\{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)=−\mathrm{1}\Rightarrow{x}=\mathrm{2}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$

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