Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 1-6 - Cumulative Review: 4

Answer

$c = \frac{e^{5} + 7}{2}$

Work Step by Step

$\ln(2c - 7) = 5$ $\log_e (2c - 7) = 5$ $e^{5} = 2c - 7$ $e^{5} + 7 = 2c$ $c = \frac{e^{5} + 7}{2}$ Check: $\ln(2(\frac{e^{5} + 7}{2}) - 7) \overset{?}{=} 5$ $\ln (e^{5} + 7 - 7) \overset{?}{=} 5$ $\ln(e^{5} ) \overset{?}{=} 5$ $5 \log_e e \overset{?}{=} 5$ $5(1) \overset{?}{=} 5$ $5 = 5$
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