Thursday, December 18, 2014

College Algebra, Chapter 5, 5.3, Section 5.3, Problem 10

Evaluate the expression log11000


log11000=log1log1000Law of Logarithms loga(AB)=logaAlogaBlog11000=0log1000Properties of Logarithms loga1=0log11000=32Because 1032=11000

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