$${{\mathtt{e}}}^{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)}{\mathtt{\,-\,}}{\mathtt{5}} = {\mathtt{8}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\frac{{ln}{\left({{\mathtt{13}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}{i}\right)}}{{ln}{\left({\mathtt{e}}\right)}}}\\
{\mathtt{x}} = {\frac{{ln}{\left({\mathtt{\,-\,}}\left({{\mathtt{13}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}\right)\right)}}{{ln}{\left({\mathtt{e}}\right)}}}\\
{\mathtt{x}} = {\frac{{ln}{\left({\mathtt{\,-\,}}\left({{\mathtt{13}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}{i}\right)\right)}}{{ln}{\left({\mathtt{e}}\right)}}}\\
{\mathtt{x}} = {\frac{{ln}{\left({\mathtt{13}}\right)}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{e}}\right)}\right)}}\\
\end{array} \right\}$$
There you go - it has got 2 real solutions and 2 imaginary solutions.
$$\\e^{4x}-5=8\\\\
e^{4x}=13\\\\
ln(e^{4x})=ln(13)\\\\
4xln(e)=ln(13)\\\\
4x=ln(13)\\\\
x=ln(13)/4$$
that is interesting, I missed out on one of the answers