If for x, y ∈ R, x > 0, y = log10 x + log10 x1/3 + log10 x1/9 + ... up to ∞ terms and 2 + 4 + 6 + ... + 2y3 + 6 + 9 + ... + 3y = 4log10 x, then the ordered pair (x, y) is equal to:

The value of 0.16log2.5(13 + 132 + 133 + ... ∞) is:

Let Sn(x) = loga1/2 x + loga1/3 x + loga1/6 x + loga1/11 x + loga1/18 x + loga1/27 x + ... up to n terms, where a > 1. If S24(x) = 1093 and S12(2x) = 265, find the value of a.