Need help in indices, exponential

given that 4^(2x-1) x 8^(1-x) = 3^(1-2x), evaluate 18^x

4^(2x-1) x 8^(1-x) = 3^(1-2x)
4^(2x) ÷ 4¹ × 8¹ ÷ 8^(x) = 3¹ ÷ 3^(2x)
4^(2x) × 2 ÷ 8^(x) = 3¹ ÷ 3^(2x)
4^(2x) ÷ 8^(x) × 3^(2x) = 3 ÷ 2
2^(4x) ÷ 2^(3x) × 3^(2x) = 3/2
2^(x) × 3^(2x) = 1½
2^(x) × 3^(x) × 3^(x) = 1½
(2×3×3)^(x) = 1½
18^(x) = 1½.

I'd make both horrible terms on the left both be in the same base. Luckily 2² = 4 and 2³=8. So:
2^2^(2x-1) * 2^3^(1-x) = 3^(1-2x)
2^2(2x-1) * 2^3(1-x) = 3^(1-2x)
2^(4x-2) * 2^(3-3x) = 3^(1-2x)
2^(4x-2+3-3x) = 3^(1-2x)
2^(x+1) = 3^(1-2x)
2^x * 2^1 = 3^1 * 3^-2x
2^x * 2 = 3 / (3^(2x))
2^x * 3^(2x) = 3/2
2^x * 3^(x+x) = 3/2
2^x * 3^x * 3^x = 3/2
(2*3*3)^x = 3/2
18^x = 3/2

4^(2x-1) x 8^(1-x) = 3^(1-2x) or
4^(2x-1) x [{4^(1-x)}*{2^(1-x)}] = 3^(1-2x)
[4^{(2x-1)}+(1-x)}]*{2^(1-x)} = 3^(1-2x) or
[4^x]*{2^(1-x)} = 3^(1-2x) or
[2^(2x)]*{2^(1-x)} = 3^(1-2x) or
[2^{(2x)+1-x}] = 3^(1-2x) or
[2^(x+1)] = 3^(1-2x) or multiplying both sides by 3^2x
(2)*(2^x)*3^(2x) = 3 or
(2^x)*{(3^2)^x} = (3/2) or
(2^x)*(9^x) = 3/2 or
18^(2x) = 3/2