Question 270807
The following table shows the amount of polonium in milligrams remaining after x days from an original sample of 2 milligrams.
x(days) 0 100 200 300
y milligrams 2 1.22 0.743 0.453
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Form: y = ab^x ; Find a and b
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2 = ab^0, so 2 = a
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Now y = 2b^x
1.22 = 2b^100
0.743 = 2b^200

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Divide the 1st into the 2nd to solve for "b"
b^100 = 0.743/1.22 = 0.6090
b = 0.6090^(1/100) = 0.9951
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Now y = 2*0.9951^x
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a. is the half life pf te polonium less than 200 days?
Let y = 1 and solve for "x":
1 = 2*(0.9951)^x
x*log(0.9951) = log(1/2)
x = 141.11 days
Ans to Question: Yes
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b. give a general formula that can be used to approximate the amount A of polonium after x days.
A(t) = 2*(0.9951)^x
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c. what is the half-life of the polonium? 
Ans: 141.11 days
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Graph A(x) and show the point that would show the half-life for x=2 milligrams.
{{{graph(400,300,-10,200,-1,3,1,2(0.9951)^x)}}}
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Cheers,
Stan H.