Question

sendo o ㏒ 2 = 0,3010 e ㏒ 3 = 0,4771 determine
a) ㏒₆ 12
b) ㏒ ₃₆ 120
c) ㏒ ₄ 30

Answer 1

A) 

Log6 base 12 

12=6^x
log12=xlog6
X=log12/log6
X= 1,0791/0,7781
X=1,3868

B) 

Log120/log36
log10.12/2log6
log10+log12/2log6
1+1,0791/.20,7781
2,0791/1,5562 = 1,3360

C) 

log30/log4 
log3.10/2log2
log3+log10/2log2
0,4771+1/2.0,3010
1,4771/0,6020 = 2,4536 


Espero ter ajudado!