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 - 5.6 Logarithmic and Exponential Equations - 5.6 Assess Your Understanding - Page 311: 45



Work Step by Step

Take the $\log_8$ of both sides: $\log_8{(8^{-x})}=\log_8{1.2}\\-x=\log_8{1.2}$ Multiply $-1$ to both sides to obtain $x=-\log_8{1.2}$ We know that $\log_a {b}=\frac{\log_c {b}}{\log_c {a}}$ (Change of Base formula). Hence, using the change-of-base formula and a calculator gives $x=-\log_8{1.2}=-\frac{\ln{1.2}}{\ln{8}}=-0.08768\approx-0.088$
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