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

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Answer 1 · 2016-03-11T02:55:25

It is

#log (1/100)=log(1/10^2)=log1-log10^2=-2log10=-2#

Note that #log_10 10=1#

Answer 2 · 2016-03-11T02:58:08

#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#