Solve $2^{2x} + 9e^{-2x} = 6$ for x using substitution. This is the equation I have:
$$2^{2x} + 9e^{-2x} = 6$$
I want to solve for x using the substitution method.
I've turned it into
$$4^x+\frac{9}{e^{2x}} - 6 = 0$$
But I do not know what to substitute and how to solve it. 
 A: Hints:
$$2^{2x}=e^{2x\log2}\implies 2^{2x} + 9e^{-2x} = 6\implies e^{2x(1+\log2)}-6e^{2x}+9=0\implies\ldots$$
Edit: Perhaps, as suggested in the comments, there's a typo in the OP, which misled me to commit a typo myself and a mistake: if the equation were
$$e^{2x}+9e^{-2x}=6\implies\left(e^{2x}\right)^2-6e^{2x}+9=0$$
and all is nice and dandy with the quadratic $\,t^2-6t+9=(t-3)^2\,$ ... As it stands now it looks like a mean, evil exponential equation of the form $\,t^{1+\log 2}-6t+9=0\,$ ...
A: This based on the equation
$$
a^b = e^{b \log a}.
$$
Apply this to the only exponential not with base $e$ in your problem:
$$
\begin{aligned}
2^{2x} + 9e^{-2x} &= 6\\
e^{2x \log 2} + 9e^{-2x} &= 6
\end{aligned}
$$
Now, the only substitution for $x$ I see is $u = e^{2x}.$
$$
\begin{aligned}
u^{\log 2} + \frac{9}{u} &= 6\\
u^{1 + \log 2} + 9 &=6u\\
u^{1 + \log 2} -6u + 9 &=0
\end{aligned}
$$
I don't see any way to solve for $u$, and so for $x$, here.  Something involving the Lambert $W$-function, perhaps.
