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La fonction Log par WinAkademy Soutien Scolaire
1. Log x +e x
v Log 1=0.
v Log e=1. (une valeur approchée de e est e=2,71828
1
v (Log x)’= ; ∀x ∈ ]0, +∞ [
x
v (Log x =Log a ) ⇔ x = a .
v (Log x p Log y) ⇔ (x p y) .
v H(x)=Log f(x) : D H = {x ∈ ¡; f (x) f 0 } ; H ‘(x) =
f '(x)
.
f (x)
v F(x) =Log f (x) : D F = {x ∈ ¡; f (x) ≠ 0 } ; F’(x) =
f '(x)
.
f (x)
v ∀a ∈ ]0, +∞ [; ∀b ∈ ]0, +∞ [: Log ab=Log a+Log b.
1
v ∀a ∈ ]0, +∞ [: Log =-Log a.
a
a
v ∀a ∈ ]0, +∞ [; ∀b ∈ ]0, +∞ [: Log =Log a –Log b.
b
v ∀r ∈ ¤ ; ∀a ∈ ]0, +∞ [: Log a r = r Log a; en particulier: ∀n ∈ ¥*
1
Log n a = Loga .
n
v ( y =e x ; x∈ ¡ ;y ∈ ]0, +∞ [) ⇔ (x = Log y ; x∈ ¡; y ∈ ]0, +∞ [).
v ∀x ∈ ¡, Log(e x ) = x .
v ∀x ∈ ]0, +∞ [ : e Logx = x.
v e 0 =1 ; e 1 = e.
b
b −
v (a Log x+b=0 ;a ≠ 0 ) ⇔ (Log x=- ) ⇔ ( x = e a
).
a
b
v (a e x +b=0 ) ⇔ (e x =- ;a ≠ 0).
a
v ∀ a ∈ ¡ , ∀b ∈ ¡ ; e a + b = e a e b .
1
v ∀a ∈ ¡ ;e − a = .
ea
ea
v ∀a ∈ ¡ ; ∀b ∈ ¡ e a − b = .
eb
v ∀ a ∈ ¡ ; ∀r ∈ ¤ , e r a =(e a ) r .
v ∀x ∈ ¡ ; (e x )’=e x .
v G(x)=e v(x) ; D G =D v et G’(x)=v’(x).e v(x) .
v (e x =e y ) ⇔ (x = y) .
v (e x p e y ) ⇔ (x p y) .