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: 30

Answer

$x=\dfrac{\ln19-1}{4}\approx0.486110$

Work Step by Step

$1+e^{4x+1}=20$ First, let's solve for $e^{4x+1}$: $e^{4x+1}=20-1$ $e^{4x+1}=19$ Apply $\ln$ to both side of the equation: $\ln e^{4x+1}=\ln19$ The exponent $4x+1$ can be taken down to multiply in front of its respective $\ln$: $(4x+1)\ln e=\ln19$ Since $\ln e=1$, the equation becomes: $4x+1=\ln19$ Solve for $x$: $4x=\ln19-1$ $x=\dfrac{\ln19-1}{4}\approx0.486110$
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