Question 979793
You might be trying to express  {{{e^(2x)-5e^x+4=0}}}.


A temporary substitution of {{{p=e^x}}} gives you  {{{p^2-5p+4=0}}}.
{{{(p-4)(p-1)=0}}}


{{{p=4}}}  or  {{{p=1}}}


Those values mean
either {{{e^x=4}}}  OR  {{{e^x=1}}}.


Solution finally is  {{{highlight(x=ln(4))}}}  or  {{{highlight(x=0)}}}.