Question 739791
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