Question Number 130553 by Study last updated on 26/Jan/21
$${log}\left({x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+….\right)=? \\ $$
Answered by Dwaipayan Shikari last updated on 26/Jan/21
$${log}\left(−{log}\left(\mathrm{1}−{x}\right)\right)={log}\left({log}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\right) \\ $$
Commented by Study last updated on 26/Jan/21
$${log}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)=? \\ $$
Commented by Study last updated on 26/Jan/21
$${why}\:{did}\:\:{you}\:{write}\:−{log}\left(\mathrm{1}−{x}\right)\:{directly}? \\ $$
Commented by mr W last updated on 27/Jan/21
$$\mathrm{ln}\:\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+… \\ $$$${replace}\:{x}\:{with}\:−{x} \\ $$$$\mathrm{ln}\:\left(\mathrm{1}−{x}\right)=−{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}−… \\ $$$$\Rightarrow−\mathrm{ln}\:\left(\mathrm{1}−{x}\right)={x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+… \\ $$$$\Rightarrow\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}={x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+… \\ $$
Commented by Study last updated on 27/Jan/21
$${log}_{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}} \left({x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+….\right)=? \\ $$