bonjour c'est un exercice sur les fonctions logarithmes quelqu'un pourrait m'aider svp ?

A=ln(2^4)-7ln(2)+4ln(2^5)+3ln(2^(-3))
  =4ln(2)-7ln(2)+20ln(2)-9ln(2)
  =8ln(2)

B=2ln(2^(3/2))-3ln(2^(-1/2))
  =3ln(2)+3/2ln(2)
  =9/2ln(2)

C=ln(1/2+√2/2)+ln(1/2-√2/2)
  =ln((1/2+√2/2)(1/2-√2/2))
  =ln(1/4-1/2)
  =ln(1/2)
  =-ln(2)

D=(ln(5)-ln(10))/(2ln(2^(1/2))
   =(ln(1/2))/(ln(2))
   =-(ln(2))/(ln(2))
   =-1

E=(1/3ln(3²)-4ln(3^(1/2)-ln(3^(-1)))/(ln(3))
  =(2/3ln(3)-2ln(3)+ln(3))/(ln(3))
  =(-1/3ln(3))/(ln(3))
  =-1/3

Bonsoir ;

[tex]\textit{On a : } 16 = 2^4 \ ;\ 32=2^5 \ ; \ \dfrac{1}{8} = \dfrac{1}{2^3} = 2^{-3} \; \\\\\\ \textit{donc on a : } A=ln(16)-7ln(2)+4ln(32)+3ln(\dfrac{1}{8}) \\\\\\ = ln(2^4) - 7ln(2) + 4ln(2^5)+3ln(2^{-3}) \\\\\\ = 4ln(2) - 7ln(2) + 20ln(2) -9ln(2) = 8ln(2) .[/tex]