Asked by Aria
                The amount of radiocarbon present after t years is given by y=ae^-.0001216t .
find the half live of 14
            
        find the half live of 14
Answers
                    Answered by
            Steve
            
    ln(1/2) = -0.69314718
So, -.0001216 = 0.00017543ln(1/2) = ln2/5700
e^(-.0001216t)
= e^(ln(1/2)*t/5700)
= (1/2)^(t/5700)
So, the half-life is 5700 years
    
So, -.0001216 = 0.00017543ln(1/2) = ln2/5700
e^(-.0001216t)
= e^(ln(1/2)*t/5700)
= (1/2)^(t/5700)
So, the half-life is 5700 years
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