Question 95229
The amount of a radioactive tracer remaining after t days is given by A=A lower case o e^-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 he oiginal amount to decay?
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A(t) =Aoe^-0.058t
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Note: If half of it decays, half of Ao remains.
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EQUATION:
(1/2)Ao = Aoe^(-0.058t)
1/2 = e^(-0,058t)
Take the natural log of both sides to get:
ln(0.5) = -0.058t
t = 11.9508 days
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Cheers,
Stan H.