Find the possible values of $x$ if $2^{2x+1} = 3(2^x) -1$ 
Find the possible values of $x$ if $2^{2x+1} = 3(2^x) -1$

I know that $x=0$ and $x=-1$ are possible values of $x$  by looking at the equation. I need help understanding how to use logarithms to solve questions of this type. Here is what I'm doing, where am I going wrong?:
$$2^{2x+1} = 3(2^x) -1$$
Can be written as $$ 3(2^x) - 2^{2x+1} =1$$
Taking logarithms of each side (and here is where I think I go wrong):
$$[x \ln(2) + \ln(3)] - [2x \ln(2) + \ln(2)] = \ln(1)$$
$$[x \ln(2) + \ln(3)] - [2x \ln(2) + \ln(2)] = 0$$
$$-x \ln(2) + \ln(3) - \ln(2) = 0$$

$$2^{2x+1} = 3(2^x) -1$$
$$2\cdot2^{2x}-3\cdot2^x+1=0$$
$2^x=t>0$
$$2t^2-3t+1=0$$
$t=1$ or $t=\frac12$
$2^x=2^{-1}$ or $2^x=1$
$x=-1$ or $x=0$
