Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Cumulative Review Exercises - Page 952: 5


The solution is, $x={{\log }_{e}}9\,\,\text{or}\,\,{{\log }_{e}}5$

Work Step by Step

Consider the given equation $\begin{align} & {{e}^{2x}}-14{{e}^{x}}+45=0 \\ & {{e}^{2x}}-9{{e}^{x}}-5{{e}^{x}}+45=0 \\ & {{e}^{x}}\left( {{e}^{x}}-9 \right)-5\left( {{e}^{x}}-9 \right)=0 \\ & \left( {{e}^{x}}-9 \right)\left( {{e}^{x}}-5 \right)=0 \end{align}$ Therefore, ${{e}^{x}}=9\,\,\text{or}\,\,5$ Taking the logarithm with base $e$ of both side, we get, $x={{\log }_{e}}9\,\,\text{or}\,\,{{\log }_{e}}5$ .
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