Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Section 3.4 - Exponential and Logarithmic Equations - Exercise Set - Page 491: 134

Answer

$ x=e^{2/0.48}-1\approx 63.5$

Work Step by Step

Solve $0.48\ln(x+1)+27=29$ (1) We graph $ g(x)=29$ in the same window and find the x-coordinate of the intersection point. The graph gives $ x=63.5.$ $(2)$ Algebraically, start by subtracting 27 from both sides. $ 0.48\ln(x+1)=2\qquad $ ... $/\div 0.48$ $\displaystyle \ln(x+1)=\frac{2}{0.48}\qquad $ ... $/e^{(...)}$ $ x+1=e^{2/0.48}$ $ x=e^{2/0.48}-1\approx 63.5000930649$ which is verified by the graphing result.
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