Question Number 107737 by mathdave last updated on 12/Aug/20
Answered by hgrocks last updated on 12/Aug/20
![ln(1−x)ln(1+x) = (1/2).((ln(1+x) + ln(1−x))^2 − ln^2 (1+x) − ln^2 (1−x)) = (1/2)[ln^2 (1−x^2 )−ln^2 (1−x)−ln^2 (1+x)] [2ab = (a+b)^(2 ) −a^2 −b^2 ] [lnx + lny = ln(xy)]](https://www.tinkutara.com/question/Q107742.png)
$$\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}.\left(\left(\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\:+\right.\right. \\ $$$$\left.\mathrm{l}\left.\mathrm{n}\left(\mathrm{1}−\mathrm{x}\right)\right)^{\mathrm{2}} \:−\:\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{x}\right)\:−\:\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)\right) \\ $$$$ \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)−\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)−\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{x}\right)\right] \\ $$$$ \\ $$$$\left[\mathrm{2ab}\:=\:\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}\:} −\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right] \\ $$$$\left[\mathrm{lnx}\:+\:\mathrm{lny}\:=\:\mathrm{ln}\left(\mathrm{xy}\right)\right] \\ $$