For this case we have an equation of the form:
y = A * (b) ^ t
Where,
A: initial amount
b: decrease rate
t: periods of time
Substituting values:
y = 64 * (0.5) ^ t
For y = 1 we have:
1 = 64 * (0.5) ^ t
Clearing t:
(0.5) ^ t = 1/64
log0.5 ((0.5) ^ t)) = log0.5 (1/64)
t = log0.5 (1/64)
t = 6
Answer:
it will be 1 gram of radioactive material left in the container after:
6 half lives
