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 - Mid-Chapter Check Point: 28

Answer

$-\dfrac{1}{2}$

Work Step by Step

Note that $0.1 = \dfrac{1}{10}$. The given expression is equivalent to: $=\log_{100}{(\frac{1}{10})}$ Note that $10=\sqrt{100} = 100^{\frac{1}{2}}$. Thus, the expression above is equivalent to: $=\log_{100}{\left(\frac{1}{100^{\frac{1}{2}}}\right)}$ Use the rule $\dfrac{1}{a^m} = a^{-m}$ to obtain: $=\log_{100}{(100^{-\frac{1}{2}})}$ RECALL: $\log_b{(b^{x})}=x$ Use the rule above with $b=100$ and $x=-\frac{1}{2}$ to obtain: $=-\dfrac{1}{2}$
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