Question 1037153
{{{e^(5x)+4e^(3x)+4e^(2x)=-9-3e^x+5e^(4x)}}}

<==> {{{e^(5x) - 5e^(4x) +4e^(3x)+4e^(2x) + 3e^x +9 = 0}}}.

Now let {{{w = e^x}}}.

Then the last equation is equivalent to 

{{{w^5 - 5w^4 +4w^3+4w^2 + 3w +9 = 0}}}.

By using the rational root theorem and trying out root candidates by synthetic division, you will find that --

{{{w^5 - 5w^4 +4w^3+4w^2 + 3w +9 = (w+1)(w-3)^2(w^2+1)}}}.

==> the zeros are w = -1, 3 (double), i, -i.

Therefore the solution set of the original equation is { ln(-1), ln3, ln3, lni, ln(-i)}
If we are after REAL zeros only, then the solution to the original equation is x = ln3, which is a double root.  ( ln(-1), lni, and ln(-i) all have values in the complex domain.)