Pythagorean comma and diesis

You can express all 5-limit intervals in terms of 25/24, 2048/2025 and
32805/32768 by setting h12, h17 and h10 to be the standard vals for
12, 17 and 10 et respectively, and then

q = (25/24)^h12(q) (2048/2025)^h17(q) (32805/32768)^h10(q)

Since 128/125 = (2048/2025) (81/80) and the Pythangorean comma is
32805/32768 times 81/80, we can express q in terms of 128/125,
3^12/2^19 and 81/80 by

q = (25/24)^h12(q) (128/125)^h17(q) (3^12/2^19)^h10(q) (81/80)^g27(q),

where g27 = h10+h17 = <27 43 63|.

If we temper the values of 25/24, 128/125 and 3^12/2^19 to what they
have in some version of meantone, and assign 81/80 to a cube root of
unity, meaning a number e not equal to 1 such that e^3=1, we get

q ~ v1^h12(q) v2^h17(q) v3^h10(q) e^g27(q)

Here in TOP, for example, v1 would be 1.04497, v2 would be 1.02403 and
v3 would be 0.97653. The exponent of e can be reduced mod 3, so we can
use the val <0 1 -1| for it. Hence 3/2 would be 697.564 cents if we
ignore the third root of unity business; to include it you might have
keyboards 0,1,and -1, and this would be 696.564 on keyboard #1. Fifths
are on keyboard 1, major thirds and fourths on keyboard -1, minor
thirds and octaves on keyboard 0, and so forth.