Saturday, October 24, 2015

College Algebra, Chapter 5, 5.3, Section 5.3, Problem 14

Evaluate the expression log3100log318log350


log3100log318log350=log3100(log318+log350)log3100log318log350=log3100log3(1850)Law of Logarithms loga(AB)=logaA+logaBlog3100log318log350=log3(100900)Law of Logarithms loga(AB)=logaAlogaBlog3100log318log350=log3(19)Simplifylog3100log318log350=2Because 32=19

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