Question 247199
{{{(4^n)^(3(2n+2)) = (4^2)^(n-1)}}}

Is that it?
If so,
3n(n+2) = 2(n-1)
3n^2 + 6n = 2n - 2
3n^2 + 4n + 2 = 0
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The solution is a complex number.
Confirm via the thank you note that it's the problem you meant.