Bonjour,
On utilise les logarithmes népériens
2,3^n > 10⁶
nln (2,3) > ln (10⁶)
n > (ln (10⁶))/(ln (2,3))
n > 16,59
→ n = 17
0,54^n < 10^(-9)
nln (0,54) < ln (10^(-9))
n >ln (10^(-9))/(ln (0,54)) car 0,54 < 1
→ tu termines ?
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Bonjour,
On utilise les logarithmes népériens
2,3^n > 10⁶
nln (2,3) > ln (10⁶)
n > (ln (10⁶))/(ln (2,3))
n > 16,59
→ n = 17
0,54^n < 10^(-9)
nln (0,54) < ln (10^(-9))
n >ln (10^(-9))/(ln (0,54)) car 0,54 < 1
→ tu termines ?