Question

Sabendo que log 3= 0,4771, log 7=0,8450 e log 11=1,0413. Calcule:
x=log 1,2...+log1,555...

Answer 1

Log 1,2.... = 
Log 12/10 = 
Log 12 - Log 10 = 
Log 12 - 1 = 
Log3 + log3+ log3 + log3 - 1 = 
0,4771 .  0,4771 .  0,4771.  0,4771 - 1 = 
1,9084 - 1 = 
0,9084

Log 1,5.... = 
Log 15/10 = 
Log15 - Log 10 = 
Log15 - 1 = 
Log(3.7-3-3) - 1 = 
(Log3 + Log7 / Log3 / Log3) -1 =
(0,4771 + 0,8450 / 0,4771/ 0,4771)-1 = 
5,8082 - 1 = 
4,8082