Question 79517
The amount of a radioactive tracer remaining after t days is given by A=Aoe^-0.058t, where Ao is the starting amount at the beginning of the time period. How many days will it take for one half of the original amount to decay?
:
Let Ao = 2; (the original amt)
Let A = 1; (half the original amt)
:
Ao*(e^-,058t) = A
:
2*(e^-.058t) = 1
:
e^-.058t = 1/2: divided both sides by 2
:
ln(e^-.058t) = ln(1/2); nat log of both sides
:
-.058t*ln(e) = ln(1/2); log equiv of exponents
:
-.058t = -.693147; remember ln(e) = 1
:
t = -.693147/-.058
:
t = +11.95 days for half of the original amt to decay.
:
:
Check on a good calc; enter: ln(e^(-.058*11.95)) = .5000