Question Number 117392 by ZiYangLee last updated on 11/Oct/20
$$\mathrm{If}\:\:\:\mathrm{log}_{\mathrm{4}} {x}+\left(\mathrm{log}_{\mathrm{4}} {x}\right)^{\mathrm{2}} +\left(\mathrm{log}_{\mathrm{4}} {x}\right)^{\mathrm{3}} +\left(\mathrm{log}_{\mathrm{4}} {x}\right)^{\mathrm{4}} +...=\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}. \\ $$
Answered by Dwaipayan Shikari last updated on 11/Oct/20
$${log}_{\mathrm{4}} {x}+\left({log}_{\mathrm{4}} {x}\right)^{\mathrm{2}} +...=\mathrm{1} \\ $$$$\frac{{log}_{\mathrm{4}} {x}}{\mathrm{1}−{log}_{\mathrm{4}} {x}}=\mathrm{1} \\ $$$$\frac{{log}_{\mathrm{4}} {x}}{{log}_{\mathrm{4}} \left(\frac{\mathrm{4}}{{x}}\right)}=\mathrm{1}\Rightarrow{log}_{\mathrm{4}} \left(\frac{\mathrm{4}}{{x}}\right)={log}_{\mathrm{4}} {x}\:\:\:\Rightarrow\frac{\mathrm{4}}{{x}}={x}\Rightarrow\:\:\:\:\:\:{x}\:=\mathrm{2} \\ $$$${So} \\ $$$${it}\:{becomes}\:{log}_{\mathrm{4}} \mathrm{2}+\left({log}_{\mathrm{4}} \mathrm{2}\right)^{\mathrm{2}} +...=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}+....=\mathrm{1} \\ $$
Answered by $@y@m last updated on 11/Oct/20
$$\frac{\mathrm{log}\:{x}}{\mathrm{1}−\mathrm{log}\:{x}}=\mathrm{1} \\ $$$$\mathrm{1}−\mathrm{log}\:{x}=\mathrm{log}\:{x} \\ $$$$\mathrm{2log}\:{x}=\mathrm{1} \\ $$$$\mathrm{log}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}=\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${x}=\mathrm{2} \\ $$
Commented by ZiYangLee last updated on 12/Oct/20
$$\mathrm{K} \\ $$