College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.4 - Logarithmic Functions - 6.4 Assess Your Understanding - Page 450: 112


$x=\ln{1.25}-1 $

Work Step by Step

Divide by 4 on both sides of the equation to obtain: $\dfrac{4e^{x+1}}{4} = \dfrac{5}{4} \\e^{x+1}=1.25$ Take the natural logarithm of both sides to obtain: $\ln{(e^{x+1})}=\ln{1.25}$ Note that $\ln{(e^x)} = x$. Thus, the equation above is equivalent to: $x+1=\ln{1.25}$ Subtract by $1$ on both sides of the equation to obtain: $x+1-1=\ln{1.25}-1 \\x=\ln{1.25}-1 $
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