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The amount of radiocarbon present after t years is given by y=ae^-.0001216t .

find the half live of 14

1 answer

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