Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.4 - Properties of Logarithms - Exercise Set - Page 713: 70

Answer

$C+A$

Work Step by Step

Note that $6=3(2)$. Thus, the given expression is equivalent to: $=\log_b{(3\cdot2)}$ RECALL: (1) $\log_b{(\frac{m}{n})} = \log_b{m} - \log_b{n}$ (2) $\log_b{mn} = \log_b{m} + \log_b{n}$ Use rule (2) above to obtain: $=\log_b{3} + \log_b{2}$ With $\log_b{2} = A$ and $\log_b{3}=C$, the expression above is equivalent to: $=C+A$
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