Question 193628
A radioactive substance that has a half-life of 8 days is to be temporarily
 implanted in a hospital patient until three-fifths of the amount originally
 present remains. How long should the implant remain in the patient?
:
The half-life equation:  Ao*(2^(-t/h)) = A
Where
A = .6; which is {{{3/5}}}
Ao = 1
h = 8 days
t = time in days
:
1*2^(-t/8) = .6
:
2^(-t/8) = .6
:
{{{(-t/8)}}}log(2) = log(.6)
find the logs of 2 and .6
{{{(-.301t/8)}}} = -.222
-.301t = 8 * -.222
-.301t = -1.7748
t = {{{(-1.7748)/(-.301)}}}
t = 5.9 days
;
:
Check on a calc: enter 2^(-5.9/8) = .599 ~ .6