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: 40

Answer

$5\log_7{x}$

Work Step by Step

RECALL: (1) $\log_a{(MN)} = \log_a{M} + \log_a{N}$ (2) $\log_a{\left(\dfrac{M}{N}\right)} = \log_a{M} - \log_a{N}$ (3) $\log_a{a} = 1$ Note that $x^5=x(x)(x)(x)(x)$. So the given expression is equivalent to: $=\log_7{\left(x\cdot x\cdot x\cdot x\cdot x\right)}$ Using rule (1) above gives: $=\log_7{x} + \log_7{x}+ \log_7{x}+ \log_7{x}+ \log_7{x}$ Factor out $\log_7{x}$ to obtain: $=\log_7{x}(1+1+1+1+1) \\=\log_7{x}(5) \\=5\log_7{x}$
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