How do you evaluate ##log (1/100)##?

##log(1/100)=-2##

First, lets assume that the base of the logarithm is 10, sometimes written ##log_(10)##. Next, we'll simplify by using the knowledge that

##log(x^a)=a*log(x)##

We can convert the ##1/100## in the expression to a power of 10:

##log(1/100)=log(100^(-1))=log((10^2)^(-1))=log(10^-2)##

Which we can rewrite as

##-2*log(10)=-2##

since ##log_10(10)=1##