Bonjour,
a)
q(0)=1000*0.5^(0.3^0)=1000*0.5*1=500
b)
1) solution à l'aide d'untableur: voir fichier joint
2) solution classique
[tex] q(t)=1000*0.5^{0.3^t}>990\\\\
ln(q(t))=ln(1000)+0.3^t* ln(0.5)>ln(990)\\\\
0.3^t * ln(0.5)>ln(990)-ln(1000)\\\\
0.3^t < \dfrac{ln(990)-ln(1000}{ln(0.5)}\\\\
0.3^t<0,01449956969511507663379792435735...\\\\
t* ln(0.3)<ln(0,01449956969511507663379792435735...)\\\\
t > \dfrac{ln(0,01449956969511507663379792435735...)}{ln(0.3)} \\\\
t > 3,5163886517895117214309871001699...\\\\
t\geq4
[/tex]
