Question Number 194165 by pascal889 last updated on 29/Jun/23
Answered by qaz last updated on 29/Jun/23
![log_2 4096+[4(log_2 x)^3 −16log_2 x]log_2 x=0 log_2 x=y ,12+4y^4 −16y^2 =0 y=±1 ,±(√3) ⇒x=2^(±1) ,2^(±(√3))](https://www.tinkutara.com/question/Q194187.png)
$${log}_{\mathrm{2}} \mathrm{4096}+\left[\mathrm{4}\left({log}_{\mathrm{2}} {x}\right)^{\mathrm{3}} −\mathrm{16}{log}_{\mathrm{2}} {x}\right]{log}_{\mathrm{2}} {x}=\mathrm{0} \\ $$$${log}_{\mathrm{2}} {x}={y}\:\:\:\:,\mathrm{12}+\mathrm{4}{y}^{\mathrm{4}} −\mathrm{16}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$${y}=\pm\mathrm{1}\:,\pm\sqrt{\mathrm{3}}\:\:\:\Rightarrow{x}=\mathrm{2}^{\pm\mathrm{1}} ,\mathrm{2}^{\pm\sqrt{\mathrm{3}}} \\ $$