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

Question

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.

Answer 1

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?

Answer 2

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 $$ \left( t-4 \right) \left( t+3 \right) \left( {t}^{2}+t+1 \right) =0$$

Answer 3

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 $$t^2-t+1=13$$

Answer 4

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?