Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises - Page 368: 29

Answer

$x=\dfrac{1-\ln12}{4}\approx-0.371227$

Work Step by Step

$8+e^{1-4x}=20$ First, let's solve for $e^{1-4x}$: $e^{1-4x}=20-8$ $e^{1-4x}=12$ Apply $\ln$ to both sides of the equation: $\ln e^{1-4x}=\ln12$ The exponent $1-4x$ can be taken down to multiply in front of its respective $\ln$: $(1-4x)\ln e=\ln12$ Since $\ln e=1$, the equation becomes: $1-4x=\ln12$ Solve for $x$: $4x=1-\ln12$ $x=\dfrac{1-\ln12}{4}\approx-0.371227$
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