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OpenStudy (anonymous):
Shoot...
OpenStudy (anonymous):
Thirty percent of a radioactive substance decays in four years. Assuming the decay is exponential, find half life of the substance.
OpenStudy (anonymous):
I am stumped. Sorry. Need to go study this one!
OpenStudy (zarkon):
pretty easy...where are you stuck
OpenStudy (anonymous):
the whole problem
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OpenStudy (anonymous):
no worry's GT
OpenStudy (zarkon):
if you start with one unit of substance , how much would you have after 4 years
OpenStudy (zarkon):
can you answer that
OpenStudy (anonymous):
no
OpenStudy (anonymous):
0.7 units
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OpenStudy (zarkon):
yes..now plug that into the exponential decay model and solve for k
\[A=A_0e^{kt}\]
OpenStudy (anonymous):
You are solving for e^kt = 1/2?
OpenStudy (zarkon):
not yet...we have to find k first
OpenStudy (anonymous):
\[\ln .70/4=k\]
OpenStudy (zarkon):
good
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OpenStudy (anonymous):
-0.089
OpenStudy (zarkon):
yes
OpenStudy (zarkon):
now use your k and solve this for t
\[\frac{1}{2}=e^{kt}\]
OpenStudy (anonymous):
@zarkon - thanks! I learned (rather refreshed my memories!) of it.
OpenStudy (zarkon):
no problem
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OpenStudy (anonymous):
ln(.5)/(?)=t
OpenStudy (anonymous):
would the -0.089 replace the ?
OpenStudy (anonymous):
It has to.
OpenStudy (zarkon):
yes...that is where k goes
OpenStudy (anonymous):
7.788=t rite
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OpenStudy (zarkon):
I get 7.7734328395
OpenStudy (zarkon):
I let the calculator keep all the digits
OpenStudy (anonymous):
ohh okay i see now
OpenStudy (anonymous):
The radioactive element polonium has a half-life of 140 days. If 15 mg of this substance decays exponentially, find:
A.) How much of the substance would remain after 60 days.
B.) how long would it take before it becomes 4 mg.
OpenStudy (zarkon):
solve
\[\frac{1}{2}=e^{k\cdot140}\]
for k
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OpenStudy (zarkon):
after finding k
A) evaluate \[15e^{k\cdot 60}\]
B) solve \[4=15e^{k\cdot t}\] for t