Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Test - Page 515: 4

Answer

$\dfrac{1}{2}$

Work Step by Step

Recall the log rules: (a) $\log_b{\dfrac{p}{q}}=\log_b{p} - \log_b{q}$ (Quotient Rule). (b) $\log{a}=\log_{10}{a}$ (c) $\log_a{a^x}=x $ We know that $36^{1/2}= \sqrt {36}=6$ Simplify the term by applying rule (c) to obtain: $\log_{36} 6 =\log_{36} 36^{1/2}=\dfrac{1}{2}$
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