Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 4 - Review - Exercises: 43

Answer

$\frac{2}{3}$

Work Step by Step

$Evaluate$ $the$ $expression$ $without$ $using$ $a$ $calculator:$ $\log_8 6 - \log_8 3 + \log_8 2$ Use the Second Law of Logarithms for $\log_8 6 - \log_8 3$ $$\log_8 6 - \log_8 3 = \log_8 (\frac{6}{3})$$ $$\log_8 2 + \log_8 2$$ Use the First Law of Logarithms $$\log_8 2 + \log_8 2 = \log_8 (2\times2)$$ $$\log_8 4$$ Use the First Law of Logarithms $$\log_8 (8\times \frac{1}{2}) = \log_8 8 + \log_8 \frac{1}{2}$$ Rewrite $\frac{1}{2}$ as $8^{-\frac{1}{3}}$ [Note: $8^{-\frac{1}{3}} = \frac{1}{8^{\frac{1}{3}}} = \frac{1}{\sqrt[3] 8} = \frac{1}{2}$] $$\log_8 8 + \log_8 8^{-\frac{1}{3}}$$ Use the Third Property of Logarithms: $\log_a a^x = x$ $$\log_8 8 + \log_8 8^{-\frac{1}{3}} = 1-\frac{1}{3}$$ $$\frac{3}{3} - \frac{1}{3} = \frac{2}{3}$$
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