Precalculus (6th Edition) Blitzer

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

Chapter 3 - Review Exercises - Page 512: 26


$-\displaystyle \frac{1}{2}$

Work Step by Step

$\log_{b}x=y $ is equivalent to $ b^{y}=x.\qquad (*)$ Consequently,$\qquad \log_{b}b^{x}=x.\qquad (**)$ --- $\displaystyle \frac{1}{\sqrt{3}}=\frac{1}{3^{1/2}}=3^{-1/2}\quad $... (we applied rules for exponents) $\displaystyle \log_{3}\frac{1}{\sqrt{3}}=\log_{3}3^{-1/2}=\quad $... apply (**) = $-\displaystyle \frac{1}{2}$
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