Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 4 - Review - Exercises - Page 389: 68



Work Step by Step

$\ln(2x-3)+1=0$ Take $1$ to the right side: $\ln(2x-3)=-1$ Rewrite this equation in exponential form: $2x-3=e^{-1}$ Solve for $x$: $2x-3=\dfrac{1}{e}$ $2x=3+\dfrac{1}{e}$ $x=\dfrac{3}{2}+\dfrac{1}{2e}\approx1.68$
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