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: 20

Answer

$2$

Work Step by Step

RECALL: (1) $\log_a{P} + \log_a{Q}=\log_a{(PQ)}$ (2) $\log_a{a} = 1$ (3) $\log_a{(a^n)}=n$ Use rule (1) above to obtain: $\log_6{9} + \log_6{4} = \log_6{(9\cdot4)}=\log_6{36}$ Note that $36=6^2$. Thus, the expression above is equivalent to: $\log_6{36}=\log_6{(6^2)}$ Use rule (3) above to obtain: $\log_6{(6^2)} = 2$
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