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: 38

Answer

$x \approx 6.06$

Work Step by Step

RECALL: $\ln{a^b} = b \cdot \ln{a}$ Take the natural logarithm of both sides to obtain: $\ln{5^{x-3}}=\ln{137}$ Use the rule above to obtain: $(x-3)\ln{5}=\ln{137}$ Divide $\ln{5}$ on both sides to obtain: $x-3= \dfrac{\ln{137}}{\ln{5}}$ Add $3$ on both sides of the equation to obtain: $x = 3 +\dfrac{\ln{137}}{\ln{5}}$ Use a scientific calculator to obtain: $x=6.056956027 \\x \approx 6.06$
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