College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.6 - Logarithmic and Exponential Equations - 6.6 Assess Your Understanding - Page 465: 45


$x = -\dfrac{\ln{1.2}}{\ln{8}} \approx -0.088$

Work Step by Step

Take the natural logarithm of both sides to obtain: $\ln{(8^{-x})} = \ln{1.2}$ Use the rule $\ln{a^x} = x \cdot \ln{a}$ to obtain: $-x\cdot \ln{8} = \ln{1.2}$ Divide both sides of the equation by $-\ln{8}$ to obtain: $\dfrac{-x\cdot \ln{8}}{-\ln{8}}=\dfrac{\ln{1.2}}{-\ln{8}} \\x = -\dfrac{\ln{1.2}}{\ln{8}} \\x \approx -0.088$
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