How do you evaluate ##log (1/100)##?
February 20th, 2023
##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##