Most drugs including colloidal silver are scavenged from the body according to their half-life. The half-life is the time required to excrete or metabolize one half of the remaining amount in the body or blood stream.

If the half-life of a drug is 1 day, and a body initially contains 1 unit of a drug, then 1 day later it will contain only half a unit, and then 1 day after that it will contain only half of a half or 1 quarter and so on.

So each day, the amount excreted decreases by 1/2 and there is only 1/2 of the previous days amount left.

To calculate the remaining amount after 3 days, we could do:

Remaining drug R = 1 -1/2 - 1/4 -1/8, which equals 1/8 left (12.5%).

An easier way is to calculate Remainder = .5^3 (.5 cubed, or .5^{3} in different notation)

So after 10 days, the remaining drug R would be .5^10 = .001 or .1% of the initial dose.

We can even calculate the remaining dose for partial half lives, like how much remains after 2 1/2 days.

R for 2.5 days = .5^2.5 or .178 (17.8%)

To calculate the remaining amount, the equation becomes:

R = .5^(N/H)

Where R = remaining fraction of initial dose

N = Time periods

H = Half Life

What if we knew the amounts and the times, and we want to calculate the half-life? Three data points are required: An initial drug level, a final drug level, and the time between the two levels. The time is the number of days or hours between the two level measurements.

To calculate the half life, we can rearrange the above equation:

Since R = .5^(N/H)

We take the logarithm of both sides giving:

log R = (N/H) * log .5

A bit of rearranging gives:

N/H = log R / log .5

Solving this equation for H gives.

H = N/(log R / log .5)

This can be rearranged to:

H = log .5 * N / log R

Since log .5 is a constant -.301:

H = -.301 * N / log R

Example 1:

Initial concentration in blood = 25mg

Final conentration in blood = 2 mg

Elapsed Time = 6 days

R = 2mg/25mg = 0.08

N = 6 days

H = -.301 * 6 * log 0.08

H = -.301 * 6 * / -1.097

H = 1.64 days half life

Example 2:

R = .125

N = 3 Days

H = -.301 * 3 / -.903 = 1 day