How do you evaluate # log 300 - log 3#?

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How do you evaluate # log 300 - log 3#?

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Answer 1 · 2015-11-04T17:02:35

Have a look:

Explanation:

I would use the property of logs that tells us:
#loga*b=loga+logb#
To get:
#log (3*100)-log(3)=#
#=log(3)+log(100)-log(3)=#
#=cancel(log3)+log(100)cancel(-log3)=#
#=log(100)#
If the base of the log is #10# then:
#log_(10)(100)=2#

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How do you evaluate # log 300 - log 3#?

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