SOLUTION: I need help. The amount of radioactive tracer remaining after t days is given by A=A0e^-0.058t, where A0 is the starting amount at the beginning of the time period. How many days

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Question 65738: I need help. The amount of radioactive tracer remaining after t days is given by A=A0e^-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?
I've had trouble with this.
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(48535) (Show Source):
You can put this solution on YOUR website!
A=A0e^-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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To check the half-life let A=(1/2)Ao
(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:
-0.6931...=-0.058t
t=11.951 years
Half-life = 11.95 yrs
Cheers,
Stan H.

Answer by josmiceli(6784) (Show Source):
You can put this solution on YOUR website!

or very close to 12 days