Question

sabendo que log 2=x e log 3=y .Calcule :

log6
log 1,5
log 2^3
log 36
log 1/9
log 16
log 2/3
log 3^log2
log√2
log 27


por favooooor,é pra amanhã !!

Answer 1

log2 = x

log3 = y

a) log6

= log(3.2)

= log3 + log2

= y + x

b) log 1,5

= log(3:2)

= log3 - log2

= y - x

c) log 2^3

= 3.log2

= 3.x

= 3x

d) log 36  

= log(3.3.2.2)

= log3 + log3 + log2 + log2

= y + y + x + x

= 2y + 2x

e) log 1/9

= log(1 : 9)

= log1 - log9

= 0 - log9

= -log9

= - log(3.3)

= - (log3 + log3)

= - log3 - log3

= -y-y

= -2y

f) log 16

= log(2^4)

= 4.log2

= 4.x

= 4x

g) log 2/3

= log (2:3)

= log2 - log3

= x - y

h) log 3^log2

= log2 . log3

= x . y

= xy

i) log√2

= log(2^(1/2))

= 1/2 . log2

= log2/2

= x/2

j) log 27

= log(3.3.3)

= log3 + log3 + log3

= y + y + y

= 3y