Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.5 - Exponential and Logarithmic Equations - Exercise Set - Page 726: 39


$x \approx 0.76$

Work Step by Step

RECALL: $\ln{a^b} = b \cdot \ln{a}$ Take the natural logarithm of both sides to obtain: $\ln{2^{x+1}}=\ln{5^x}$ Use the rule above to obtain: $(x+1)\ln{2}=x\ln{5}$ Distribute $\ln{2}$ to obtain: $x\ln{2} + \ln{2} = x\ln5$ Subtract $\ln{2}$ on both sides to obtain: $x\ln{2} = x\ln{5}-\ln{2}$ Subtract $x\ln{5}$ on both sides of the equation to obtain: $x\ln{2} - x\ln{5}=-\ln{2}$ Factor out $x$ on the left side of the equation to obtain: $x(\ln2-\ln5)=-\ln{2}$ Divide $(\ln{2} - \ln{5})$ on both sides of the equation to obtain: $x = \dfrac{-\ln{2}}{\ln{2}-\ln{5}}$ Use a scientific calculator to obtain: $x= 0.7564707974 \\x \approx 0.76$
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