College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.5 - Properties of Logarithms - 6.5 Assess Your Understanding: 24

Answer

$2$

Work Step by Step

RECALL: (1)The change-of-base formula for logarithms: $\log_a{b} = \dfrac{\log{b}}{\log{a}}$ (2) $\log_a{(B^n)}=n\cdot\log_a{B}$ Use rule (1) above to obtain: $\log_3{8} \cdot \log_8{9}=\dfrac{\log{8}}{\log{3}} \cdot \dfrac{\log{9}}{\log{8}}$ Cancel the common factors to obtain: $\require{cancel} =\dfrac{\cancel{\log{8}}}{\log{3}} \cdot \dfrac{\log{9}}{\cancel{\log{8}}} \\=\dfrac{\log{9}}{\log{3}} \\=\dfrac{\log{3^2}}{\log{3}}$ Use rule (2) above to obtain: $=\dfrac{2\log{3}}{\log{3}}$ Cancel the common factors to obtain: $\require{cancel} =\dfrac{2\cancel{\log{3}}}{\cancel{\log{3}}} \\=2$
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