<PRE><B>Question 73053</B>
The amount of a radioactive tracer remaining after t days is given by A = AO e-0.18t, where A0 is the starting amount at the beginning of the time period. How much should be acquired now to have 40 grams remaining after 3 days?
:
Radio active decay formula should be given as:
A = Ao(e^-.18t)
:
A = 40; t = 3; Find Ao
: 
40 = Ao(e^-.18*3)
40 = Ao(e^-.54
:
Using a good calculator; enter e^(-.54), should equal .5827482524
So you have:
.5827482524Ao = 40
:
Ao = 40/.5827482524
Ao = 68.64 grams to be acquired
:
If they want you to use natural logs, you add logs when you multiply:
ln(40) = ln(Ao) + ln(e^-.54)
ln(40) = ln(Ao) + (-.54)(ln(e), log equiv of exponents
:
Find the ln of 40. Remember the ln of e is 1, so we have:
3.688879454 = ln(Ao) - .54
:
3.688879454 +.54 = ln(Ao)
4.228879454 = ln(Ao)
:
Find the e^x; enter e^4.228879454:
Ao = 68.64, the same answer

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