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: 27

Answer

$-2$

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}{e^{2}}=e^{-2}\quad $... (we applied a rule for exponents) The natural logarithm is logarithm with base $ e,\quad(\ln x=\log_{e}x).$ $\displaystyle \ln\frac{1}{e^{2}}=\log_{e}e^{-2}=\quad $... apply (**) = $-2$
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