Answer
The value of x is $\left\{ \ln 2,\ln 4 \right\}$.
Work Step by Step
Consider the expression ${{e}^{2x}}-6{{e}^{x}}+8=0$
Simplify ${{e}^{2x}}$
${{e}^{2x}}={{\left( {{e}^{x}} \right)}^{2}}$
Now, take ${{e}^{x}}=u $
So, the equation will be:
${{u}^{2}}-6u+8=0$
Factorize the above equation as follows:
$\begin{align}
& {{u}^{2}}-6u+8=0 \\
& {{u}^{2}}-4u-2u+8=0 \\
& u\left( u-4 \right)-2\left( u-4 \right)=0 \\
& \left( u-4 \right)\left( u-2 \right)=0
\end{align}$
So, the factors are:
$\begin{align}
& \left( u-4 \right)=0 \\
& u=4
\end{align}$
Or,
$\begin{align}
& \left( u-2 \right)=0 \\
& u=2
\end{align}$
Now, substitute the value of $ u=4$ in equation ${{e}^{x}}=u $:
$\begin{align}
& {{e}^{x}}=u \\
& {{e}^{x}}=4 \\
\end{align}$
Take natural log on both sides;
$\begin{align}
& \ln \left( {{e}^{x}} \right)=\ln \left( 4 \right) \\
& x=\ln \left( 4 \right)
\end{align}$
Now, substitute the value of $ u=2$ in equation ${{e}^{x}}=u $:
$\begin{align}
& {{e}^{x}}=u \\
& {{e}^{x}}=2 \\
\end{align}$
Take natural log on both sides;
$\begin{align}
& \ln \left( {{e}^{x}} \right)=\ln \left( 2 \right) \\
& x=\ln \left( 2 \right)
\end{align}$
Hence, the value of x in the expression ${{e}^{2x}}-6{{e}^{x}}+8=0$ is $\left\{ \ln 2,\ln 4 \right\}$.
