e^(2x+1) = 32^6x...Solve for x?

e^(2x + 1) = 32^(6x)
ln[e^(2x + 1)] = ln[32^(6x)]
(2x + 1)ln(e) = (6x)ln(32)
(2x + 1)(1) = (6x)ln(32)
2x + 1 = (6x)ln(32)
2x - (6x)ln(32) = -1
x[2 - 6ln(32)] = -1
x = -1/[2 - 6ln(32)]

2x + 1 = (6x) ln 32
1 = (6x) ln 32 - 2x
1 = 2x [ 3 ln 32 - 1 ]
x = 1 / [ 2 (3 ln 32 - 1) ]
x = 0.146