SOLUTION: 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

Question 95229This question is from textbook Algebra and Trigonometry
: 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? This question is from textbook Algebra and Trigonometry

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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
============
Cheers,
Stan H.
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