Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Review - Exercises - Page 389: 48

Answer

$\log 4 + 3\log x - 2\log y + 5\log (x-1)$

Work Step by Step

$Expand$ $the$ $logarithmic$ $expression:$ $\log (\frac{4x^3}{y^2(x-1)^5})$ Use the Second Law of Logarithms $\log (\frac{4x^3}{y^2(x-1)^5}) = \log 4x^3 - \log y^2(x-1)^5$ Use the First Law of Logarithms for $\log 4x^3$ and $\log y^2(x-1)^5$ $\log (4\times x^3) = \log 4 + \log x^3$ $\log (y^2(x-1)^5) = \log y^2 + \log (x-1)^5$ $\log 4x^3 - \log y^2 + \log (x-1)^5$ Use the Third Law of Logarithms for the whole expression $\log 4 + \log x^3 - \log y^2 + \log (x-1)^5 = \log 4 + 3\log x - 2\log y + 5\log (x-1)$
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