ln(a b) is equal to

Rewriting this using logs instead of exponents, we see that ln (a · b) = m + n = lna + lnb (vi) If, in (v), instead of multiplying we divide, that is a b

lna = lnb est équivalent à a = b • lna < lnb Par conséquent : ln(ax) = ln(a)+ln(x) (le logarithme d'un produit est égal à la somme des logarithmes) Propriétés

the function F(x) which we define to be equal to the definite integral ln(ax) = lna + lnx, or, replacing x by b, we get the familiar identity for logarithms: ln(ab) 

Since both functions have equal derivatives, f (x) + C = g(x) for (iii) ln(a b ) = ln a − ln b ▻ Note that 0 = ln 1 = ln a a = ln(a · 1 a ) = ln a + ln 1 a , giving us 

Théorème 1 Pour tous réels a et b stritement positifs : lnab = lna + lnb lna + ln 1 b= lna ¡ lnb Corollaire 2 Si a1;a2;:::;an sont n réels strictement positifs :

Notice, however, that if n < 0 then this only has domain equal to R − {0} So we can think of (b) ln(ab) = ln(a) + ln(b) for positive numbers a and b (c) ln(a b ) 

On la note lna La fonction a) x = ea est équivalent à a = lnx avec x > 0 b) ln1= 0 ; lne = 1 ; ln 1 e x = eln y ⇔ ln x = ln y b) x < y ⇔ eln x < eln y ⇔ ln x < ln y 

b = lna − lnb 7 lna r = r lna One important application of these properties is We use the fact that if two quantities are equal, the logarithm of these quantities

It follows that eln y = ex ln b But, since eln y = y = bx , it follows that Using the formula ax = ex ln a, write 5x as ex ln 5 We can then apply the Chain Rule, 

Logarithme et Exponentielle : eln x = ln(ex) = x ln 1 = 0 ln(ab) = ln(a) + ln(b) ln(a/b ) = ln(a) − ln(b) ln(1/a) = − ln(a) ln( √a) = ln(a)/2 ln(aα) = α ln(a) e0 = 1 ex+y =