Solving $e^{4x}+3e^{2x}-28=0$ How to solve this equation:

$$e^{4x}+3e^{2x}-28=0$$

I don't know how to solve this problem. I read over another example, $e^{2x}-2e^x-8=0,$ and it said that $e^{2x}$ is $e$ to the $x$ squared, so $e^{2x}=e^{{x}^{2}}$. Why is this so? So $e^{4x}=e^{{x}^{4}}$?  
Please help me to solve it. 
 A: Hint:
$$\eqalign{e^{4x}+3e^{2x}-28=0&\implies (e^{2x})^2+3(e^{2x})-28=0\\&\implies e^{2x}=\ldots\\&\implies x\:\:\:=\ldots}$$
And I think what "$e$ to the $x$ squared" means is $(e^{x})^2$, in this case it is equal to $e^{2x}$ since one have $(a^{m})^n=a^{mn}.$
A: $$
\Big(e^{2x}\Big)^2 + 3e^{2x}-28=0
$$
$$
u^2 + 3u - 28=0
$$
This is a quadratic equation.  Once you've found $u$, you've got $e^{2x}$.  After that, find $x$.
A: Well, $e^{2x}=(e^x)^2$, but neither is equal to the "$e^{x^2}$" that you mentioned. 
So no, $e^{4x} \not= e^{x^4}$, but rather $e^{4x}=(e^4)^x$.
Therefore, $e^{4x}+3e^{2x}-28=0$ means $(e^{2x})^2+3(e^{2x})-28=0$. 
Let $m=e^{2x}$, and you well get an auxiliary trinomial of $$m^2+3m-28=0.$$ If you factor your auxiliary trinomial, you will have $$(m+7)(m-4)=0.$$ Given the roots of $m=-7$ and $m=4$, can you solve for $x$ from here?
A: $$(e^{2x})^2+3(e^{2x})-28=0.$$
$$e^{2x}=y$$
$$y^2+3y-28=0$$
$$y_{1,2}=\frac{-3\pm11}{2}$$
$$y_1=4=e^{2x},2x=\ln 4$$
$$x=\ln2$$
for $$y_2=-7$$ no real solutions 
