How do you evaluate #log 0.001 + log 100#?

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Answer 1 · 2016-07-07T04:33:33

#log 0.001 + log 100 = -1#

The common logarithm #log(x)# and exponentiation #10^x# are essentially inverse functions of one another.

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#