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 - Page 459: 22

Answer

$1$

Work Step by Step

RECALL: (1) $\log_a{P} - \log_a{Q}=\log_a{\left(\frac{P}{Q}\right)}$ (2) $\log_a{a} = 1$ (3) $\log_a{(a^n)}=n$ Use rule (1) above to obtain: $\log_8{16} - \log_8{2} = \log_8{(\frac{16}{2})}=\log_8{8}$ Use rule (2) above to obtain: $\log_8{8} = 1$
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