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Question-91608

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Question Number 91608 by mhmd last updated on 01/May/20
Commented by Tony Lin last updated on 01/May/20
let log_2 x=t (1/4)t^2 −t+4=0 t^2 −4t+16=0 t=((4±48i)/2)=2±24i log_2 x=2±24i x=2^(2±24i) =4×2^(±24i) =4×e^(±24ln2i) =4×[cos(24ln2)±isin(24ln2)] ≈4×(−0.6±0.8i) =−2.4±3.2i
$${let}\:{log}_{\mathrm{2}} {x}={t} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}{t}^{\mathrm{2}} −{t}+\mathrm{4}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{16}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{4}\pm\mathrm{48}{i}}{\mathrm{2}}=\mathrm{2}\pm\mathrm{24}{i} \\ $$$${log}_{\mathrm{2}} {x}=\mathrm{2}\pm\mathrm{24}{i} \\ $$$${x}=\mathrm{2}^{\mathrm{2}\pm\mathrm{24}{i}} =\mathrm{4}×\mathrm{2}^{\pm\mathrm{24}{i}} \\ $$$$=\mathrm{4}×{e}^{\pm\mathrm{24}{ln}\mathrm{2}{i}} \\ $$$$=\mathrm{4}×\left[{cos}\left(\mathrm{24}{ln}\mathrm{2}\right)\pm{isin}\left(\mathrm{24}{ln}\mathrm{2}\right)\right] \\ $$$$\approx\mathrm{4}×\left(−\mathrm{0}.\mathrm{6}\pm\mathrm{0}.\mathrm{8}{i}\right) \\ $$$$=−\mathrm{2}.\mathrm{4}\pm\mathrm{3}.\mathrm{2}{i} \\ $$
Commented by abdomathmax last updated on 02/May/20
(e)→(((ln(x))/(2ln2)))^2 −((ln(x))/(ln(2))) +4 =0 ⇒ ((ln^2 (x))/(4ln^2 (2)))−((lnx)/(ln(2))) +4 =0 ⇒ 4ln^2 (2){ ((ln^2 (x))/(4ln^(2)) 2))−((lnx)/(ln2)) +4}=0 ⇒ ln^2 (x)−4ln(2)ln(x) +16ln^2 (2) =0 let ln(x)=t ⇒t^2 −4ln(2)t +16ln^2 (2) =0 Δ^′ =4ln^2 2−16ln^2 2 =−12ln^2 (2) =(2i(√3)ln(2))^2 t_1 =2ln(2)+2i(√3)ln(2) =2ln(2)(1+i(√3)) =4ln(2)e^((iπ)/3) and t_2 =4ln(2)e^(−((iπ)/3)) x_1 =e^t_1 =e^(2ln(2)) e^(i(2(√3)ln2)) =4(cos(2(√3)ln(2))+isin(2(√3)ln(2)) x_2 =4{cos(2(√3)ln2)−i sin(2(√3)ln(2)}
$$\left({e}\right)\rightarrow\left(\frac{{ln}\left({x}\right)}{\mathrm{2}{ln}\mathrm{2}}\right)^{\mathrm{2}} −\frac{{ln}\left({x}\right)}{{ln}\left(\mathrm{2}\right)}\:+\mathrm{4}\:=\mathrm{0}\:\Rightarrow \\ $$$$\frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{4}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}−\frac{{lnx}}{{ln}\left(\mathrm{2}\right)}\:+\mathrm{4}\:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{4}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\left\{\:\frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{4}{ln}^{\left.\mathrm{2}\right)} \mathrm{2}}−\frac{{lnx}}{{ln}\mathrm{2}}\:+\mathrm{4}\right\}=\mathrm{0}\:\Rightarrow \\ $$$${ln}^{\mathrm{2}} \left({x}\right)−\mathrm{4}{ln}\left(\mathrm{2}\right){ln}\left({x}\right)\:+\mathrm{16}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\:=\mathrm{0} \\ $$$${let}\:{ln}\left({x}\right)={t}\:\:\Rightarrow{t}^{\mathrm{2}} −\mathrm{4}{ln}\left(\mathrm{2}\right){t}\:+\mathrm{16}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\:=\mathrm{0} \\ $$$$\Delta^{'} \:=\mathrm{4}{ln}^{\mathrm{2}} \mathrm{2}−\mathrm{16}{ln}^{\mathrm{2}} \mathrm{2}\:=−\mathrm{12}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\:=\left(\mathrm{2}{i}\sqrt{\mathrm{3}}{ln}\left(\mathrm{2}\right)\right)^{\mathrm{2}} \\ $$$${t}_{\mathrm{1}} =\mathrm{2}{ln}\left(\mathrm{2}\right)+\mathrm{2}{i}\sqrt{\mathrm{3}}{ln}\left(\mathrm{2}\right)\:=\mathrm{2}{ln}\left(\mathrm{2}\right)\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\right) \\ $$$$=\mathrm{4}{ln}\left(\mathrm{2}\right){e}^{\frac{{i}\pi}{\mathrm{3}}} \:{and}\:{t}_{\mathrm{2}} =\mathrm{4}{ln}\left(\mathrm{2}\right){e}^{−\frac{{i}\pi}{\mathrm{3}}} \\ $$$${x}_{\mathrm{1}} ={e}^{{t}_{\mathrm{1}} } \:\:={e}^{\mathrm{2}{ln}\left(\mathrm{2}\right)} \:{e}^{{i}\left(\mathrm{2}\sqrt{\mathrm{3}}{ln}\mathrm{2}\right)} \\ $$$$=\mathrm{4}\left({cos}\left(\mathrm{2}\sqrt{\mathrm{3}}{ln}\left(\mathrm{2}\right)\right)+{isin}\left(\mathrm{2}\sqrt{\mathrm{3}}{ln}\left(\mathrm{2}\right)\right)\right. \\ $$$${x}_{\mathrm{2}} =\mathrm{4}\left\{{cos}\left(\mathrm{2}\sqrt{\mathrm{3}}{ln}\mathrm{2}\right)−{i}\:{sin}\left(\mathrm{2}\sqrt{\mathrm{3}}{ln}\left(\mathrm{2}\right)\right\}\right. \\ $$$$ \\ $$

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