2^ 2x+1 = 3(2^x) - 1 = 0 how to solve this

Do i let

2^2x = y

so it would be:

y^2 = 3y - 1?

2^(2x + 1) = 3*(2^x) - 1

2*2^(2x) - 3*(2^x) + 1= 0

2*2^(2x) - 2*(2^x) - (2^x) + 1 = 0

(2*2^(x) - 1)(2^x - 1) = 0

2^x = 1/2 --> x = -1

2^x = 1 --> x = 0

use log on both sides

basically, 2^(2x+1) = 0
so 4^x = 0
hence when taking log, you can not solve as log(0) is undefined.

for 3(2^x) - 1 = 0
therefore 3(2^x) -= 1
and so 2^x = 1/3

now take log of both side gives

x log(2) = log(1/3)

x = log(1/3)/log(2)