SOLUTION: Use the exponential decay function P(t) = Poe^kt Chemistry. The decay rate of krypton-85 is 6.3% per day. What is the half-life?

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Question 173403This question is from textbook introduction to college mathematics
: Use the exponential decay function P(t) = Poe^kt
Chemistry. The decay rate of krypton-85 is 6.3% per day. What is the half-life? This question is from textbook introduction to college mathematics

Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
the formula is:
p%5Bt%5D+=+p%5B0%5D%2Ae%5E%28-k%2At%29
that minus k is important since the exponent has to be negative in a decay problem.
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you are given:
decay rate is 6.3% per day.
this translates to .063 per day which is equal to k since k, in this case, is the decay rate per day.
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you are also asked for the half life.
this means at what point in time will the amount of krypton be half of what it is today.
if p%5B0%5D = 1 = the amount of krypton today, then p%5Bt%5D = .5 is the amount of krypton when one half life has occurred.
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your equation:
p%5Bt%5D+=+p%5B0%5D%2Ae%5E%28-k%2At%29
becomes:
.5+=+1+%2A+e%5E%28-.063%2At%29
which becomes:
.5+=+e%5E%28-.063%2At%29
which you can solve using the log or ln function of your calculator.
we'll use the log function.
either one will get the same answer.
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your equation becomes:
log%2810%2C.5%29+=+log%2810%2Ce%5E%28-.063%2At%29%29
since log%2810%2Cb%5Ea%29+=+a%2Alog%2810%2Cb%29 your equation becomes:
log%2810%2C.5%29+=+-.063%2At%2Alog%2810%2Ce%29
if you divide both sides of this equation by -.063%2Alog%2810%2Ce%29, you get:
log%2810%2C.5%29%2F%28-.063%2Alog%2810%2Ce%29%29+=+t
which you can solve using your calculator to get:
t = 11.0023362 days
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to prove this is true, plug that value in your original equation for t.
that equation:
p%5Bt%5D+=+p%5B0%5D%2Ae%5E%28-k%2At%29
becomes:
.5+=+1%2Ae%5E%28-.063%2A11.0023362%29
which becomes:
.5+=+e%5E%28-.693145181%29
which becomes:
.5 = .5
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value is good.
answer is:
t = 11.0023362