How do you evaluate #\log 14+ \log 2- \log 4#?

1 Answer
hev1
Apr 8, 2018

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

Using the above logarithm rules:
#log14 + log2 - log4#
#=log((14*2)/4)#
#= log(28/4)#
#=log7#
#~~0.845#

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