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: 46


$x = -\dfrac{\ln{1.5}}{\ln{2}} \approx -0.585$

Work Step by Step

Take the natural logarithm of both sides to obtain: $\ln{(2^{-x})} = \ln{1.5}$ Use the rule $\ln{a^x} = x \cdot \ln{a}$ to obtain: $-x\cdot \ln{2} = \ln{1.5}$ Divide both sides of the equation by $-\ln{2}$ to obtain: $\dfrac{-x\cdot \ln{2}}{-\ln{2}}=\dfrac{\ln{1.5}}{-\ln{2}} \\x = -\dfrac{\ln{1.5}}{\ln{2}} \\x \approx -0.585$
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