How do you simplify #log 4 + log 5 - log 2#?

To answer this question, the following logarithm rules are needed:
#loga+logb = log(a*b)#
#loga - logb = log(a/b)#
#log_b b = 1#

Using the above rules:
#log4 + log5 - log2#
#=log((4*5)/2)#
#= log(20/2)#
#= log(10)#
Since the base of a normal #log# is #10#, #log10 = 1# (come #10^1 = 10#).