TITLE: How to solve $4^x - 2^{x + 1} = 3 $ for x?
QUESTION [8 upvotes]: We figured that this can be changed to $2^{2x} - 2^x \cdot 2 = 3$, but couldn't solve from there.  Perhaps we are not on the right path?

REPLY [7 votes]: You are on the right path.
Substitute $y=2^x$, you get $y^2-2y-3=0$. Try to solve that and change it back to $x$.

REPLY [4 votes]: $4^x - 2^{x + 1} = 3$
$$(2^{x})^2 - 2\cdot2^{x}-3 =0,\qquad 2^x=t$$
$$t^2-2t-3=0$$
$$t_{1,2}=3,-1$$
$$2^x=3\Rightarrow x=\log_2 3$$