Log a x log b x logb a where we can choose b to be

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Unformatted text preview: hm and so Property 7 can’t be used here. We do, however, have a product inside the logarithm so we can use Property 5 on this logarithm. log 4 ( x3 y 5 ) = log 4 ( x3 ) + log 4 ( y 5 ) Now that we’ve done this we can use Property 7 on each of these individual logarithms to get the final simplified answer. log 4 ( x3 y 5 ) = 3log 4 x + 5log 4 y [Return to Problems] æ x9 y5 ö (b) log ç 3 ÷ èz ø In this case we’ve got a product and a quotient in the logarithm. In these cases it is almost always best to deal with the quotient before dealing with the product. Here is the first step in this part. æ x9 y5 ö log ç 3 ÷ = log ( x 9 y 5 ) - log z 3 èz ø Now, we’ll break up the product in the first term and once we’ve done that we’ll take care of the exponents on the terms. © 2007 Paul Dawkins 291 http://tutorial.math.lamar.edu/terms.aspx College Algebra æ x9 y5 ö log ç 3 ÷ = log ( x 9 y 5 ) - log z 3 èz ø = log x 9 + log y 5 - log z 3 = 9 log x + 5log y - 3log z [Return to Problems]...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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