College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Chapter 4 Review - Concept Check - Page 425: 68

Answer

$x\approx 1.68$

Work Step by Step

Subtract $1$ on both sides of the equation to obtain: $\ln{(2x-3)}=-1$ RECALL: $\ln{x} = y \longrightarrow e^y=x$ Use the rule above to obtain: $e^{-1} = 2x-3$ Use the rule $a^{-m} = \dfrac{1}{a^m}$ to obtain: $\dfrac{1}{e} = 2x-3$ Add $3$ on both sides of the equation to obtain: $\dfrac{1}{e} + 3 = 2x$ Divide $2$ on both sides of the equation to obtain: $\dfrac{\frac{1}{e} + 3}{2}=x$ Use a calculator to obtain: $x\approx 1.68$
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