How do you evaluate #log 100#?

#log100=2#

To evaluate this you use the definition of a logarithm.
#log_ab=c iff a^c=b#

You also have to assume that if no base #b# is written then the base is #10#

So in the example you have #log_(10)100=c iff 10^c=100#.

Now you can easily find that #c=2#, because #10^2=10*10=100#.

So the answer is
#log100=log_(10)100=2#

#log100=2# because it can be denoted as

#log100 = log(10 * 10) = log10+log10#

Keep in mind that when there is no base written, it is assumed to be a base of #10#. Applying the rule of #log x=1#, you have #log10=1#.

Then that means

#log10+log10=1 + 1 = 2#

Also, you can use the power rule.

#log 100 = log(10^2) =2*log10#

which is equal to

#2*1=2#

It might help to compare the two notations:

Index form and log form.#" "# They are interchangeable.

#a^b = c hArr loga_c =b#

Index form makes a statement:

#10^1 = 10," " 10^2 = 100," "10^3 = 1000#

Log form asks a question.....

#log_10 100 =???#

"What power of #10# will give #100#?

#log_10 100 =2#

In the same way..

#log_4 16 = 2" because " 4^2 = 16#

#log_3 27 = 3" because " 3^3=27#

#log_2 32 = 5" because " 2^5=32#