SOLUTION: PLEASE HELP ME WITH THIS PROBLEM I CANNOT FIGURE IT OUT IVE BEEN TRYING FOR DAYS: The amount of a radioactive tracer remaining after t days is given A=Ao e^-0.058t, where Ao is th

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Question 55170This question is from textbook TRIGONOMETRY
: PLEASE HELP ME WITH THIS PROBLEM I CANNOT FIGURE IT OUT IVE BEEN TRYING FOR DAYS:
The amount of a radioactive tracer remaining after t days is given A=Ao 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 the original amount to decay? This question is from textbook TRIGONOMETRY

Found 2 solutions by stanbon, Hook:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A=Ao 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 the original amount to decay?
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If half of it decays that leave A=(1/2)Ao, so:
(1/2)Ao = Aoe^(-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= 11.95 days
Cheers,
Stan H.

Answer by Hook(36) (Show Source):
You can put this solution on YOUR website!
You've got the following formula and you want to know when half of the crap decays. Ok...so what that says in math-speak is:

Let's plug that into the original equation:

The Ao terms cancel out of each side, leaving

You just have to solve for t.