dx / dt = -kx where k is a constant.

dx / x = -k dt

Integrating:

log(x) – log(x0) = -kt

where x0 is the amount of the single dose.

log(x / x0) = -kt

x / x0 = e^(-kt) …(1)

As x / x0 = 0.6 when t = 1, (1) gives:

e^(-k) = 0.6 …(2)

Applying this for all the doses taken at the start of each of the first n days:

If s is the total remaining n days after commencement, then:

s = x0[ e^(-kn) + e^(-k(n – 1)) + … + e^(-k)]

This is a geometric series of n terms with first term e^(-kn) and common ratio e^(k).

s = x0[ e^(-kn) (e^(kn) – 1) ] / (e^k – 1)

s = x0 e^(-k) [ 1 – e^(-kn) ] / (1 – e^(-k)) …(2)

Putting x0 = 50mg, k = -log(0.6), n = 3 in (2):

s = 50 * 0.6 * ( 1 – 0.6^3) / (1 – 0.6)

= 58.8mg.

Putting n = x instead of 3:

s = 50 * 0.6 * ( 1 – 0.6^x) / (1 – 0.6)

= 75(1 – 0.6^x) mg.