College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Chapter 4 Review - Concept Check: 54

Answer

$\log_5{\left(\dfrac{2x+2}{(3x+7)^{\frac{1}{3}}}\right)}$

Work Step by Step

RECALL: (1) $n \cdot \log_a{P}=\log_a{(P^n)}$ (2) $\log_a{P} + \log_a{Q}=\log_a{(PQ)}$ (3) $\log_a{P} - \log_a{Q}=\log_a{(\frac{P}{Q})}$ (4) $a^m \cdot a^n=a^{m+n}$ Use rule (1) above to obtain $\log_5{2}+\log_5{(x+1)}-\frac{1}{3}\log_5{(3x+7)} \\=\log_5{2}+\log_5{(x+1)}-\log_5{(3x+7)^{\frac{1}{3}}}$ Use rule (2) above to obtain: $\log_5{2}+\log_5{(x+1)}-\log_5{(3x+7)^{\frac{1}{3}}} \\=\log_5{[2(x+1)]}-\log_5{(3x+7)^{\frac{1}{3}}} \\=\log_5{(2x+2)}-\log_5{(3x+7)^{\frac{1}{3}}}$ Use rule (3) above to obtain: $\log_5{(2x+2)}-\log_5{(3x+7)^{\frac{1}{3}}} \\=\log_5{\left(\dfrac{2x+2}{(3x+7)^{\frac{1}{3}}}\right)}$
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