Bonjour Alisson,
Rappels :
log(a*b) = log(a) + log(b)
log(a/b) = log(a) - log(b)
log() = b*log(a)
log(10) = 1
1)
a. On commence par faire apparaitre les 3 et 5.
log(5*9) = log(5*3*3)
On utlise la première formule : log(a*b) = log(a)+log(b)
donc log(5*3*3) = log(5) + log(3) + log(3)
= log(5) + 2log(3)
b. log(a/b) = log(a) - log(b)
donc log(5/9) = log(5) - log(9)
= log(5) - log(3*3)
= log(5) - 2log(3)
c. log() = b*log(a)
donc log() = 3*log(5)
2) On essaie a chaque fois de se ramener à du log(10)
a. log() = 5log(10)
or log(10) = 1
donc 5log(10) = 5
b. log( ) = -9log(10)
donc -9log(10) = -9
c. log( ) = log() - log( )
= 3log(10) - (-2)log(10)
= 3log(10) + 2log(10)
= 5log(10)
d. log() = log(1) = 0
Bonne journée
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Bonjour Alisson,
Rappels :
log(a*b) = log(a) + log(b)
log(a/b) = log(a) - log(b)
log() = b*log(a)
log(10) = 1
1)
a. On commence par faire apparaitre les 3 et 5.
log(5*9) = log(5*3*3)
On utlise la première formule : log(a*b) = log(a)+log(b)
donc log(5*3*3) = log(5) + log(3) + log(3)
= log(5) + 2log(3)
b. log(a/b) = log(a) - log(b)
donc log(5/9) = log(5) - log(9)
= log(5) - log(3*3)
= log(5) - 2log(3)
c. log() = b*log(a)
donc log() = 3*log(5)
2) On essaie a chaque fois de se ramener à du log(10)
a. log() = 5log(10)
or log(10) = 1
donc 5log(10) = 5
b. log( ) = -9log(10)
or log(10) = 1
donc -9log(10) = -9
c. log( ) = log() - log( )
= 3log(10) - (-2)log(10)
= 3log(10) + 2log(10)
= 5log(10)
or log(10) = 1
donc 5log(10) = 5
d. log() = log(1) = 0
Bonne journée