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: 40

Answer

$x \approx 1.71$

Work Step by Step

RECALL: $\ln{a^b} = b \cdot \ln{a}$ Take the natural logarithm of both sides to obtain: $\ln{4^{x+1}}=\ln{9^x}$ Use the rule above to obtain: $(x+1)\ln{4}=x\ln{9}$ Distribute $\ln{4}$ to obtain: $x\ln{4} + \ln{4} = x\ln9$ Subtract $\ln{4}$ on both sides to obtain: $x\ln{4} = x\ln{9}-\ln{4}$ Subtract $x\ln{9}$ on both sides of the equation to obtain: $x\ln{4} - x\ln{9}=-\ln{4}$ Factor out $x$ on the left side of the equation to obtain: $x(\ln4-\ln9)=-\ln{4}$ Divide $(\ln{4} - \ln{9})$ on both sides of the equation to obtain: $x = \dfrac{-\ln{4}}{\ln{4}-\ln{9}}$ Use a scientific calculator to obtain: $x= 1.709511291 \\x \approx 1.71$
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