Question

considerando log 2=0,4771,calcule:
log 12
log48
log64

Answer 1

 Considerando log 2=0,4771,calcule:

log 12 ==> log(2² . 3) ==> log2² + log3
 2log2 +log3
2.(0,477) + log3 ==> 0,954 + log3

log48 ==> log(2^4 . 3) ==> 4log2 + log3
4.(0,477) + log3 ==> 1,908 + log3

log64 ==>  log(2^6) ==> 6log2
 6.(0,477) ==> 2,862

Answer 2

a) $log12 =$

$log (3 * 4) =$

$log (3 * 2^2)=$

$log3 + log2^2$ 

$log3 +2log2=$ 

$0,4771 +2*0,301=$ 

$1,0791$ 

b) $log48 =$

$log(3*16)$ =

$log3+log16$ =

$log3 + log2^4=$ 

$log3 + 4log2=$ 

$0,4771 + 4*0,301=$ 

$1,6811$

c) $log64=$

$log(4 *16)=$

$log4+log16=$

$log2^2+log2^4=$

$2*0,301+4*0,301 =$

$1,806$

Espero ter ajudado.