${\left( {e}^{x} \right)}^{3}-7{e}^{x}+6=0$
To get an equation that is easier to solve, substitute $t$ for ${e}^{x}$${t}^{3}-7t+6=0$
Solve the equation for $t$$\begin{array} { l }t=1,\\t=-3,\\t=2\end{array}$
Substitute back $t={e}^{x}$$\begin{array} { l }{e}^{x}=1,\\{e}^{x}=-3,\\{e}^{x}=2\end{array}$
Solve the equation for $x$$\begin{array} { l }x=0,\\{e}^{x}=-3,\\{e}^{x}=2\end{array}$
Solve the equation for $x$$\begin{array} { l }x=0,\\x\notin ℝ,\\{e}^{x}=2\end{array}$
Solve the equation for $x$$\begin{array} { l }x=0,\\x\notin ℝ,\\x=\ln\left({2}\right)\end{array}$
Find the union$\begin{array} { l }x=0,\\x=\ln\left({2}\right)\end{array}$
The equation has $2$ solutions$\begin{align*}&\begin{array} { l }x_1=0,& x_2=\ln\left({2}\right)\end{array} \\&\begin{array} { l }x_1=0,& x_2\approx0.693147\end{array}\end{align*}$