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: 21

Answer

$\dfrac{1}{2}$

Work Step by Step

Note that $10=\sqrt{100}$. Replace $10$ with its equivalent of $\sqrt{100}$ to obtain: $=\log_{100}{(\sqrt{100})}$ Use the rule $\sqrt{a}= a^{\frac{1}{2}}$ to obtain: $=\log_{100}{(100^{\frac{1}{2}})}$ RECALL: $\log_b{b^x} = x$ Use the rule above to obtain: $=\dfrac{1}{2}$
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