SOLUTION: 6. Given log(3) = 0.47712 , find a approximation for each logarithm a) log(30) b) log(3000) c) log(.3) d) log(.003) e) log(9) f) log(81) g) log(√3) Can you please h

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: 6. Given log(3) = 0.47712 , find a approximation for each logarithm a) log(30) b) log(3000) c) log(.3) d) log(.003) e) log(9) f) log(81) g) log(√3) Can you please h      Log On


   

Question 739791: 6. Given log(3) = 0.47712 , find a approximation for each logarithm
a) log(30)
b) log(3000)
c) log(.3)
d) log(.003)
e) log(9)
f) log(81)
g) log(√3)
Can you please help me out? Thanks so much in advance:)
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
6. Given log(3) = 0.47712 , find a approximation for each logarithm
a) log(30)=log(10*3)=log(10)+log(3)=1+.47712=1.47712
b) log(3000)=log(1000*3)=log(1000)+log(3)=3+.47712=3.47712
c) log(.3)=log(10^(-1)*3)=-1+log(3)=-1+.47712=-0.52288
d) log(.003)=log(10^(-3)*3)=-3+log(3)=-3+.47712=-2.52288
e) log(9)=log(3^2)=2log(3)=2*.47712=0.95424
f) log(81)=log(3^3)=3log(3)=3*.47712=1.43136
g) log(√3)=(1/2)log(3)=1/2*.47712=0.23856
note: log of base raised to an exponent=exponent; e.g. log 10^3=3