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: 24

Answer

$x \approx 1.69$

Work Step by Step

RECALL: $\log_b{a^x}=x \cdot \log_b{a}$ Take the natural logarithm of both sides to obtain: $\ln{19^x}=\ln{143}$ Use the rule above with $b=e$ to obtain: $x \cdot \ln{19}=\ln{143}$ Divide $\ln{19}$ to both sides of the equation to obtain: $x =\dfrac{\ln{143}}{\ln{19}}$ Use a scientific calculator to obtain: $x=1.685497531 \\x \approx 1.69$
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