How do you solve ${e^{2x}} - 6{e^x} + 8 = 0$?

Hint: Here the given equation can be solved by two different methods. One we can directly split the middle term or we can substitute any variable in the equation and then frame the simplified equation and then split the middle term and find the resultant values for “x”.

Complete step-by-step solution:
Take the given equation: ${e^{2x}} - 6{e^x} + 8 = 0$
The above equation can be re-written as:
${\left( {{e^x}} \right)^2} - 6{e^x} + 8 = 0$
We can factorize the above equation without substituting.
${\left( {{e^x}} \right)^2} - 4{e^x} - 2{e^x} + 8 = 0$
Make a pair of first two terms and the last two terms.
$\underline {{{\left( {{e^x}} \right)}^2} - 4{e^x}} - \underline {2{e^x} + 8} = 0$
Take out common factors from the paired terms.
\[{e^x}({e^x} - 4) - 2({e^x} - 4) = 0\]
Take multiple common from the above equation.
\[({e^x} - 4)({e^x} - 2) = 0\]
Above equation implies,
\[{e^x} = 4\] or \[{e^x} = 2\]
Take natural log on both sides of the equation.
\[\ln {e^x} = \ln 4\] or \[\ln {e^x} = \ln 2\]
Natural log and exponential cancel each other on the left hand side of the equation.
$x = \ln 4$or $x = \ln 2$
Second Alternative method:
Here we will assume one unknown term for ${e^x}$
Let, $t = {e^x}$ …. (A)
Substitute in the given equation.
$ \Rightarrow {t^2} - 6t + 8 = 0$
The above equation can be solved splitting the middle term.
$ \Rightarrow {t^2} - 4t - 2t + 8 = 0$
Make a pair of first two terms and the last two terms.
$ \Rightarrow \underline {{t^2} - 4t} - \underline {2t + 8} = 0$
Take out common factors from the paired terms.
\[ \Rightarrow t(t - 4) - 2(t - 4) = 0\]
Take multiple common from the above equation.
\[ \Rightarrow (t - 4)(t - 2) = 0\]
Above equation implies,
\[ \Rightarrow t = 4\;{\text{or t = 2}}\]
Replace the values from equation (A)
\[{e^x} = 4\] or \[{e^x} = 2\]
Take natural log on both sides of the equation.
\[\ln {e^x} = \ln 4\] or \[\ln {e^x} = \ln 2\]
Natural log and exponential cancel each other on the left hand side of the equation.
$x = \ln 4$or $x = \ln 2$
This is the required solution.

Note: Be careful about the sign convention. Always remember when we have to simplify between the two negative signs then you have to do addition in actual and then place the sign of negative, such as $ - 2 - 6 = - 8$