Question 72299
The amount of a radioactive tracer remaining after t days is given
by A = A0 e–0.058t, where A0 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?
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If half decays that leaves (1/2)Ao.
EQUATION:
(1/2)Ao = Ao e^(-0.058t)
Divide both sides by Ao to get:
1/2 = e^(-0.058t)
Take the natural log of both sides to get:
ln(1/2) = -0.058t
t= ln*1/2) / -0.058 
t=11.95 days
The half-life of the tracer is 11.95 days.
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Cheers,
Stan H.