Answer
$x = ln(9)$
$x = ln(5)$
Work Step by Step
$$e^{2x} - 14e^{x} + 45 = 0$$
This equation can be re-written as follows:
$$(e^{x})^{2} - 14(e^{x}) + 45 = 0$$
where we can, initially, substitute the value ($e^{x}$) for an arbitrary variable $y$ and end with a quadratic equation:
$$y^{2} - 14y + 45 = 0$$
This equation can be easily factorized since two factors of $45$ that, when added, yield $-14$ are ($-9\times-5$):
$$(y-9)(y-5) = 0$$
This results in $y=9$ and $y=5$. Since we established earlier that $y = e^{x}$, we can now solve for the original variable $x$:
$$(e^{x}) = 9$$
$$(e^{x}) = 5$$
By converting these equations from their exponential form to their logarithmic form, we conclude that:
$$(e^{x}) = 9$$
$log_{e}9 = ln9 = x$
$$(e^{x}) = 5$$
$log_{e}5 = ln5 = x$
