Question Number 171031 by cortano1 last updated on 06/Jun/22
Answered by a.lgnaoui last updated on 07/Jun/22
![=Log((a/(1−b)))+Log(((1−b)/(1+b)))=Log((a/(1−b))×((1−b)/(1+b)))=Log((a/(1+b))) =Log(((a−b+1)/(a+b+1)))^3 =3Log(((a+1−b)/(a+1+b)))=3Log((((a+1)^2 −b^2 )/((a+1+b)^2 ))) =3Log[(((a+1)^2 −(1−a^2 ))/((a+1)^2 +b^2 +2b(a+1)))]=3Log[(((a+1)((a+1)−(1−a)))/((a^2 +2a+1+2(a+1)b+b^2 ))]= =3Log[((2a(a+1))/(2+2a+2(a+1)b))]=3Log[((2a(a+1))/(2(a+1)(1+b)))]= =3Log((a/(1+b))) donc ((Log((a/(1−b)))+Log(((1−b)/(1+b))))/(Log(((a−b+1)/(a+b+1)))^3 ))=((Log((a/(b+1))))/(3Log((a/(b+1))))) =(1/3)](https://www.tinkutara.com/question/Q171054.png)
$$={Log}\left(\frac{{a}}{\mathrm{1}−{b}}\right)+{Log}\left(\frac{\mathrm{1}−{b}}{\mathrm{1}+{b}}\right)={Log}\left(\frac{{a}}{\mathrm{1}−{b}}×\frac{\mathrm{1}−{b}}{\mathrm{1}+{b}}\right)={Log}\left(\frac{{a}}{\mathrm{1}+{b}}\right) \\ $$$$={Log}\left(\frac{{a}−{b}+\mathrm{1}}{{a}+{b}+\mathrm{1}}\right)^{\mathrm{3}} =\mathrm{3}{Log}\left(\frac{{a}+\mathrm{1}−{b}}{{a}+\mathrm{1}+{b}}\right)=\mathrm{3}{Log}\left(\frac{\left({a}+\mathrm{1}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }{\left({a}+\mathrm{1}+{b}\right)^{\mathrm{2}} }\right) \\ $$$$=\mathrm{3}{Log}\left[\frac{\left({a}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{b}\left({a}+\mathrm{1}\right)}\right]=\mathrm{3}{Log}\left[\frac{\left({a}+\mathrm{1}\right)\left(\left({a}+\mathrm{1}\right)−\left(\mathrm{1}−{a}\right)\right)}{\left({a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}+\mathrm{2}\left({a}+\mathrm{1}\right){b}+{b}^{\mathrm{2}} \right.}\right]= \\ $$$$=\mathrm{3}{Log}\left[\frac{\mathrm{2}{a}\left({a}+\mathrm{1}\right)}{\mathrm{2}+\mathrm{2}{a}+\mathrm{2}\left({a}+\mathrm{1}\right){b}}\right]=\mathrm{3}{Log}\left[\frac{\mathrm{2}{a}\left({a}+\mathrm{1}\right)}{\mathrm{2}\left({a}+\mathrm{1}\right)\left(\mathrm{1}+{b}\right)}\right]= \\ $$$$=\mathrm{3}{Log}\left(\frac{{a}}{\mathrm{1}+{b}}\right) \\ $$$$\:\:{donc}\:\:\:\frac{{Log}\left(\frac{{a}}{\mathrm{1}−{b}}\right)+{Log}\left(\frac{\mathrm{1}−{b}}{\mathrm{1}+{b}}\right)}{{Log}\left(\frac{{a}−{b}+\mathrm{1}}{{a}+{b}+\mathrm{1}}\right)^{\mathrm{3}} }=\frac{{Log}\left(\frac{{a}}{{b}+\mathrm{1}}\right)}{\mathrm{3}{Log}\left(\frac{{a}}{{b}+\mathrm{1}}\right)}\:\:=\frac{\mathrm{1}}{\mathrm{3}}\:\:\: \\ $$