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: 24

Answer

$\frac{1}{2}\log_5{x} - 2$

Work Step by Step

RECALL: (1) $\log{(b^c)}=c \cdot \log{b}$ (2) $\log_b{(xy)} = \log_b{x} + \log_b{y}$ (3) $\log_b{(\frac{x}{y})}=\log_b{x} - \log_b{y}$ Use rule (3) above to obtain: $=log_5{\sqrt{x}}-\log_5{25}$ Note that $\sqrt{x} = x^{\frac{1}{2}}$ and $25=5^2$. Thus, the expression above is equivalent to: $=\log_5{(x^{\frac{1}{2}})}-\log_5{(5^2)}$ Use rule (1) above to obtain: $=\frac{1}{2}\log_5{x} - 2\log_5{5}$ Use the rule $\log_b{b} = 1$ to obtain: $=\frac{1}{2}\log_5{x} - 2 \cdot 1 \\=\frac{1}{2}\log_5{x} - 2 $
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