Answer
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,
$x={{\log }_{e}}9\,\,\text{or}\,\,{{\log }_{e}}5$
.
