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Question Number 154989 by bagjagugum123 last updated on 24/Sep/21

x_1 and x_2 is root log_2 x^((1+^2 log x)) =2, the value is x_1 +x_2 = ... a. 2(1/4) b. 2(1/2) c. 4(1/4) d. 4(1/2) e. 6(1/4)

$${x}_{\mathrm{1}} \:{and}\:{x}_{\mathrm{2}} \:{is}\:{root}\:\mathrm{log}_{\mathrm{2}} {x}^{\left(\mathrm{1}+^{\mathrm{2}} \mathrm{log}\:{x}\right)} =\mathrm{2},\:{the}\:{value} \\ $$$${is}\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\:... \\ $$$${a}.\:\mathrm{2}\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${b}.\:\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${c}.\:\mathrm{4}\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${d}.\:\mathrm{4}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${e}.\:\mathrm{6}\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Answered by john_santu last updated on 24/Sep/21

⇔ (log _2 x+1)log _2 x = 2 let y=log _2 x ⇒y^2 +y−2=0 ⇒(y+2)(y−1)=0 { ((y=−2⇒x_1 =(1/4))),((y=1⇒x_2 =2)) :} ⇒x_1 +x_2 =(9/4)

$$\Leftrightarrow\:\left(\mathrm{log}\:_{\mathrm{2}} {x}+\mathrm{1}\right)\mathrm{log}\:_{\mathrm{2}} {x}\:=\:\mathrm{2} \\ $$$${let}\:{y}=\mathrm{log}\:_{\mathrm{2}} {x} \\ $$$$\Rightarrow{y}^{\mathrm{2}} +{y}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left({y}+\mathrm{2}\right)\left({y}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\begin{cases}{{y}=−\mathrm{2}\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}}\\{{y}=\mathrm{1}\Rightarrow{x}_{\mathrm{2}} =\mathrm{2}}\end{cases}\:\Rightarrow{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}} \\ $$

Commented by bagjagugum123 last updated on 24/Sep/21

thank you very much Sir

$${thank}\:{you}\:{very}\:{much}\:{Sir} \\ $$

Answered by puissant last updated on 24/Sep/21

⇒ (1+log_2 x)log_2 x=2 t=log_2 x → t^2 +t−2=0 ⇒ t_1 =−2 ; t_2 =1 ((lnx_1 )/(ln2))=−2 ⇒ lnx=−2ln2 ⇒ x=(1/4) ((lnx_2 )/(ln2))=1⇒ lnx_2 =ln2 ⇒ x_2 =2 ∴∵ x_1 +x_2 =(1/4)+(8/4)=(9/4)..

$$\Rightarrow\:\left(\mathrm{1}+{log}_{\mathrm{2}} {x}\right){log}_{\mathrm{2}} {x}=\mathrm{2} \\ $$$${t}={log}_{\mathrm{2}} {x}\:\rightarrow\:{t}^{\mathrm{2}} +{t}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\:{t}_{\mathrm{1}} =−\mathrm{2}\:\:;\:\:{t}_{\mathrm{2}} =\mathrm{1} \\ $$$$\frac{{lnx}_{\mathrm{1}} }{{ln}\mathrm{2}}=−\mathrm{2}\:\Rightarrow\:{lnx}=−\mathrm{2}{ln}\mathrm{2}\:\Rightarrow\:{x}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{{lnx}_{\mathrm{2}} }{{ln}\mathrm{2}}=\mathrm{1}\Rightarrow\:{lnx}_{\mathrm{2}} ={ln}\mathrm{2}\:\Rightarrow\:{x}_{\mathrm{2}} =\mathrm{2} \\ $$$$\therefore\because\:\:\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{8}}{\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{4}}.. \\ $$

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