Solve $\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}=13$ for $x$ 
Solve $$\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}=13$$ for $x$ 

Any help is appreciated. I'm entering a challenge and can't reach the solution.
 A: write your equation in the form
$$\frac{2^{4x}+2^{2x}+1}{2^{2x}+2^x+1}=13$$
Setting $$t=2^x$$ you will get
$$t^4+t^2+1=13(t^2+t+1)$$
factorizing the whole equation we get
$$ \left( t-4 \right)  \left( t+3 \right)  \left( {t}^{2}+t+1 \right) =0$$
A: Let $u=2^x$.  After converting $4^{2x}$ to $2^{4x}=u^4$, etc., the equation to solve becomes
$${u^4+u^2+1\over u^2+u+1}=13$$
Can you take it from there?
A: Alternative approach: notice that
$$t^4+t^2+1 = (t^2+1)^2-t^2 = (t^2+t+1)(t^2-t+1)$$
so
$$\frac{t^4+t^2+1}{t^2+t+1} = t^2-t+1$$
so either
$$t^2+t+1=0$$
or
$$t^2-t+1=13$$
A: First, we can rearrange the given equation as follows:
\begin{align}\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}&=13\\\\
\frac{(2^2)^{2x}+(2^2)^x+1}{2^{2x}+2^x+1}&=13\\\\
(2^2)^{2x}+(2^2)^x+1&=13\left(2^{2x}+2^x+1\right)\\
2^{4x}+2^{2x}+1&=13\left(2^{2x}+2^x+1\right)\\
(2^x)^4+(2^x)^2+1&=13\left((2^x)^2+2^x+1\right)\\
y^4+y^2+1&=13\left(y^2+y+1\right)\tag{substitute $y:=2^x$}\\
y^4+y^2+1&=13y^2+13y+13\\
y^4-12y^2-13y-12&=0\end{align}
We are then left with a quartic equation to solve, and then we can solve $y=2^x$ for each of the values of $y$ we find. 
Can you continue from here?
