How do you evaluate #10 log18 - log3#?

1 Answer
Alan P.
Dec 21, 2015

#10log(18)-log(3)=log(18^10/3)#

Explanation:

Remember

  1. #mlog(n) = log(n^m)#
  2. #log(a)-log(b) = log(a/b)#

Therefore
#color(white)("XXX")10log(18)-log(3)#

#color(white)("XXX")=log(18^10)-log(3)#

#color(white)("XXX")=log(18^10/3)#

This could be evaluated using a calculator as:
#color(white)("XXX")=12.0756#

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