## College Algebra 7th Edition

Published by Brooks Cole

# Chapter 4, Exponential and Logarithmic Functions - Chapter 4 Review - Concept Check - Page 425: 43

#### Answer

$\dfrac{2}{3}$

#### Work Step by Step

RECALL: (1) $\log_a{P} - \log_a{Q} = \log_a{(\frac{P}{Q})}$ (2) $\log_a{P} + \log_a{Q} = \log_a{(PQ)}$ Use rule (1) above to obtain: $=\log_8{(\frac{6}{3})} + \log_8{2} \\=\log_8{2} + \log_8{2}$ Use rule (2) above to obtain: $=\log_8{(2\cdot2)} \\=\log_8{4}$ RECALL: $\log_a{b} = \dfrac{\log{b}}{\log{a}}$ Use the rule above to obtain: $\log_8{4} = \dfrac{\log{4}}{\log{8}}$ Since $4=2^2$ and $8=2^3$, then the expression above is equivalent to: $=\dfrac{\log{(2^2)}}{\log{(2^3)}}$ RECALL: $\log{(M^r)}=r \cdot \log{M}$ Use the rule above to obtain: $\require{cancel} \dfrac{\log{(2^2)}}{\log{(2^3)}} \\=\dfrac{2\log{2}}{3\log{2}} \\=\dfrac{2\cancel{\log{2}}}{3\cancel{\log{2}}} \\=\dfrac{2}{3}$

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