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 - Page 712: 29


$1 + \frac{1}{2}\log{x}$

Work Step by Step

Simplify the radical to obtain: $=\sqrt{10^2\cdot x} = 10\sqrt{x}$ Thus, the given expression is equivalent to: $=\log{(10\sqrt{x})}$ 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 (2) above to obtain: $=\log{10}+\log{(\sqrt{x})}$ Note that $\sqrt{x} = x^{\frac{1}{2}}$ therefore the expression above is equivalent to: $=\log{10} + \log{(x^{\frac{1}{2}})}$ Use rule (1) above to obtain: $=\log{10} + \frac{1}{2}\log{x}$ Use the rule $\log{10} = 1$ to obtain: $=1 + \frac{1}{2}\log{x}$
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