How do you evaluate #log 0.001 + log 100#?
1 Answer
Jul 7, 2016
#log 0.001 + log 100 = -1#
Explanation:
The common logarithm
So we find:
#log 0.001 + log 100 = log(10^(-3))+log(10^2) = -3+2 = -1#
Note also that in general we have:
#log(a) + log(b) = log(ab)#
So:
#log 0.001 + log 100 = log (0.001*100) = log(0.1) = log(10^(-1)) = -1#