\[{2^x + 2^{x+1} + 2^{x+2} + \ldots + 2^{x+2015} = 4^x + 4^{x+1} + 4^{x+2} + \ldots + 4^{x+2015}}\]

If \(x\) satisfies the equation above and it can be represented as \(\log_D \left(\dfrac{A}{1+B^C} \right)\) for positive integers \(A\), \(B\), \(C\), and \(D\), where \(B\) is prime, determine the smallest value of \(A + B + C +D\).

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