Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 4 - Section 4.4 - Laws of Logarithms - 4.4 exercises - Page 359: 43

Answer

$\ln x$ + $\frac{1}{2}$$($$\ln y$ - $\ln z$$)$

Work Step by Step

$Expand$ $the$ $expression$: $\ln (x\sqrt{\frac{y}{z}})$ Apply the First Law of Logarithms $\ln (x\times \sqrt{\frac{y}{z}})$ = $\ln x$ + $\ln\sqrt{\frac{y}{z}}$ Rewrite the square root for $\sqrt{\frac{y}{z}}$ $\ln x$ + $ln (\frac{y}{z})^\frac{1}{2}$ Apply the Third Law of Logarithms for $ln (\frac{y}{z})^\frac{1}{2}$ $ln (\frac{y}{z})^\frac{1}{2}$ = $\frac{1}{2}$$\ln \frac{y}{z}$ Apply the Second Law of Logarithms for $\frac{1}{2}$$\ln \frac{y}{z}$ (Distribute the half) $\frac{1}{2}$$\ln (\frac{y}{z})$ = $\frac{1}{2}$$\ln y$ - $\frac{1}{2}$$\ln z$ Assemble the expression $\ln x$ + $\frac{1}{2}$$\ln y$ - $\frac{1}{2}$$\ln z$ Factor out the $\frac{1}{2}$. We do this since one may put together the $\ln x$ and the $\frac{1}{2}$$\ln y$ when we combine it to one expression $\ln x$ + $\frac{1}{2}$$($$\ln y$ - $\ln z$$)$
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