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: 34

Answer

$\dfrac{3}{2}$

Work Step by Step

Note that $8=2^3$. Thus, $\log_4{8}=\log_4{(2^3)}$ RECALL: (1) $\log_a{b} = \dfrac{\log{b}}{log{a}}$ (2) $\log_a{(b^n)}=n \cdot \log_a{b}$ Use rule (1) above to obtain: $\log_4{(2^3)} = \dfrac{\log{(2^3)}}{\log{4}}$ With $4=2^2$, the expression above is equivalent to: $=\dfrac{\log{(2^3)}}{\log{(2^2)}}$ Use rule (2) above to obtain: $\require{cancel} \dfrac{\log{(2^3)}}{\log{(2^2)}} \\=\dfrac{3\log{2}}{2\log{2}} \\=\dfrac{3\cancel{\log{2}}}{2\cancel{\log{2}}} \\=\dfrac{3}{2}$
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