Fonction logarithme népérien, cours de Terminale - MathsFG

f f(ax) = f(a) + f(x)
a x
fRa x
f(ax) = f(a) + f(x)f0(x) = k
xk
f f(1) = f(1 ×1) = f(1) + f(1)
f(1) = 2f(1) f(1) = 0 a > 0x > 0f(ax) =
f(a) + f(x)af0(ax) = f0(x)x= 1 af0(a) = f0(1) f0(a) = f0(1)
a
x > 0f0(x) = f0(1)
xf0(1) = k k
f0(1) f0(x) = 1
xx > 0
ln ]0; +∞[
ln0(x) = 1
xln(1) = 0
]0; +∞[
xln0(x) = 1
x>0
limx7→+∞ln(x) = +∞limx7→0+ln(x) = −∞
x= 0

x+∞
ln0(x)k
k+∞
ln(x)k %
k −∞
x+∞
ln(x)k
0 1 2 3 4 5−1 0
1
2
3
−1
−2
−3
−4
0 1 2 3 4 5−1 0
1
2
3
−1
−2
−3
−4
e
ln(e) = 1 e≈2,72
]0; +∞[
u
Iln(u)I
(ln(u))0=u0
u

a b ln(ab) = ln(a) + ln(b)
a > 0x > 0h(x) = ln(ax)−ln(a)−ln(x)h
h]0; +∞[x > 0h0(x) = a1
ax −1 x= 0 h0
h h(1) = ln(a)−ln(a)−ln(1) = 0 h
x > 0a > 0 ln(ax) = ln(a) + ln(x)
a b
•ln(a
b) = ln(a)−ln(b)
•nln(an) = nln a
•ln(√a) = 1
2ln(a)
•ln(b×1
b) = ln(1) = 0
ln(b×1
b) = ln(b) + ln(1 b)
ln(b) + ln(1
b)=0
ln(a
b) = ln(a×1 b) = ln(a) + ln(1 b) = ln(a)−ln(b)
•ln(an) = ln(a) + ln(a) + . . . + ln(a)
| {z }
n
•ln(a) = ln(√a×√a) = ln(√a) + ln(√a) = 2 ln(√a)
•ln(32) = ln(25) = 5 ln(2)
•ln(1/108) = −8 ln(10)
ln(a) = ln(b)a=b
]0; +∞[

•ln(x−1) = 0
x−1>0x > 1 ]1; +∞[
ln(x−1) = 0 ln(x−1) = ln(1) x−1 = 1 x= 2 2
•ln(x−1) = 2
]1; +∞[
ln(x−1) = 2 ln(x−1) = ln(e2)x−1 = e2x= 1+e2
1 + e2
ln(a)≥ln(b)a≥b
]0; +∞[
ln(2x−1) <5
2x−1>0x > 1
2]1 2; +∞[
ln(2x−1) <5 ln(2x−1) <5 ln(e) ln(2x−1) <ln(e5)
2x−1< e5x < e5+1
2
e5+1
2
u0
uln |u|
u