SOLUTION: 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

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?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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

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