what is 4(3^2x) = e^x

Take logs of both sides and use the properties of logarithms (see the Formulary):

4*3^2x = e^x take logs (use natural log):

ln(4)+2x*ln(3) = x Rearrange and collect terms in x:

(2*ln(3)-1)*x = -ln(4) so:

x = -ln(4)/(2*ln(3)-1)

$${\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{ln}{\left({\mathtt{4}}\right)}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{3}}\right)}{\mathtt{\,-\,}}{\mathtt{1}}\right)}} = {\mathtt{x}} = -{\mathtt{1.157\: \!923\: \!406\: \!654\: \!701\: \!8}}$$