Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.4 - Properties of Logarithms - Exercise Set: 74

Answer

$\frac{1}{2}C - 2A$

Work Step by Step

Note that $16=2^4$. Thus, the given expression is equivalent to: $=\log_b{\sqrt{\left(\frac{3}{2^4}\right)}}$ Use the rule $\sqrt{m} = m^{\frac{1}{2}}$ to obtain: $= \log_b{\left[\left(\frac{3}{2^4}\right)^{\frac{1}{2}}\right]}$ RECALL: (1) $\log_b{(\frac{m}{n})} = \log_b{m} - \log_b{n}$ (2) $\log_b{mn} = \log_b{m} + \log_b{n}$ (3) $\log_b{(m^n)} = n \cdot \log_b{m}$ Use rule (3) above to obtain: $=\frac{1}{2} \cdot \log_b{\left(\frac{3}{2^4}\right)}$ Use rule (1) to obtain: $=\frac{1}{2}\left(\log_b{3} - \log_b{(2^4)}\right)$ Use rule (3) to obtain: $=\frac{1}{2}\left(\log_b{3} - 4\log_b{2}\right)$ With $\log_b{2} = A$ and $\log_b{3}=C$, the expression above is equivalent to: $=\frac{1}{2}(C-4A) \\=\frac{1}{2}C - 2A$
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