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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)

=Log(a1−b)+Log(1−b1+b)=Log(a1−b×1−b1+b)=Log(a1+b)=Log(a−b+1a+b+1)3=3Log(a+1−ba+1+b)=3Log((a+1)2−b2(a+1+b)2)=3Log[(a+1)2−(1−a2)(a+1)2+b2+2b(a+1)]=3Log[(a+1)((a+1)−(1−a))(a2+2a+1+2(a+1)b+b2]==3Log[2a(a+1)2+2a+2(a+1)b]=3Log[2a(a+1)2(a+1)(1+b)]==3Log(a1+b)doncLog(a1−b)+Log(1−b1+b)Log(a−b+1a+b+1)3=Log(ab+1)3Log(ab+1)=13

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