Alg_Complete

# The real issue here is that we cant write 8 as a

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Unformatted text preview: log y 6 We now have a difference of two logarithms and so we can use Property 6 in reverse. When using Property 6 in reverse remember that the term from the logarithm that is subtracted off goes in the denominator of the quotient. Here is the answer to this part. æ x3 ö 3log x - 6 log y = log ç 6 ÷ èy ø [Return to Problems] (c) In this case we’ve got three terms to deal with and none of the properties have three terms in them. That isn’t a problem. Let’s first take care of the coefficients and at the same time we’ll factor a minus sign out of the last two terms. The reason for this will be apparent in the next step. 5ln ( x + y ) - 2 ln y - 8ln x = ln ( x + y ) - ( ln y 2 + ln x8 ) 5 Now, notice that the quantity in the parenthesis is a sum of two logarithms and so can be combined into a single logarithm with a product as follows, © 2007 Paul Dawkins 293 http://tutorial.math.lamar.edu/terms.aspx College Algebra 5ln ( x + y ) - 2 ln y - 8ln x = ln ( x + y ) - ln ( y 2 x8 ) 5 Now we are down to two logarithms and they are a diff...
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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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