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: 37

Answer

$x \approx 1.09$

Work Step by Step

RECALL: $\ln{a^b} = b \cdot \ln{a}$ Take the natural logarithm of both sides to obtain: $\ln{7^{x+2}}=\ln{410}$ Use the rule above to obtain: $(x+2)\ln{7}=\ln{410}$ Divide $\ln{7}$ on both sides to obtain: $x+2= \dfrac{\ln{410}}{\ln{7}}$ Subtract $2$ on both sides of the equation to obtain: $x = -2 +\dfrac{\ln{410}}{\ln{7}}$ Use a scientific calculator to obtain: $x=1.091693192 \\x \approx 1.09$
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