Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.5 Properties of Logarithms - 5.5 Assess Your Understanding - Page 305: 27

Answer

$4$

Work Step by Step

We know that $\log_a {x^n}=n\cdot \log_a {x}$. Hence, $e^{\log_{e^2}{16}}=e^{\log_{e^2}{4^2}}=e^{2\log_{e^2}{4}}$. We also know that $a^{\log_a {x}}=x$. Thus, $e^{2\log_{e^2}{4}} \\=(e^2)^{\log_{e^2}{4}} \\=4$
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