If a substance has a half life of 850 years, and there are 35 milligrams present now, how long will it take for the substance to be reduced to 10 miligrams calculus

This is a problem involving differential equation. The rate of decaying substance is directly proportional to the amount of substance.
dP/dt = kA
where k = proportionality constant, A = amount of subtance.

Integrating that equation yields,
A = ce^(kt)
where c = constant of integration.
The constant of integration is represented mostly as an initial amount of substance.

We have given two initial condition:
(1) t = 0, A = 35 milligrams (present)
(2) t = 850 years, A = 35/2 milligrams (half-life)

We are going to solve for c:
35 = ce^0
c = 35

Next, solving for k:
35/2 = 35e^(k*850)
1/2 = e^(k*850)
Multiplying natural logarithm on both sides to bring down k:
ln (1/2) = k*850
k = ln (1/2) / 850
k = -0.00081547

Finally, the time when the amount of substance has 10 milligrams remaining is:
10 = 35e^(-0.00081547*t)
10/35 = e^(-0.00081547*t)
Multiplying natural logarithm on both sides to bring down t:
ln (2/7) = -0.00081547*t
t = ln (2/7) / -0.00081547
t = 1536.25 years (ANSWER)