That function may be used to model radioactive decay. Q represents the quantity remaining after t years. K is the decay constant for plutonium-240. What is the decay constant if its half life is 6300 years?

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That function may be used to model radioactive decay. Q represents the quantity remaining after t years. K is the decay constant for plutonium-240. What is the decay constant if its half life is 6300 years?

Mathematics
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\[Q(t) = Q_0 e^{-kt}\] the half-life is 6300 years, so we know the following: \[Q(0) = Q_0\]\[Q(6300) = \frac{1}2 Q_0\]because\(Q_0\) represents the initial quantity, and after one half-life, there is only half as much left. That means that we can find the value of \(k\) by setting \[Q(6300) = \frac{1}{2} Q_0 = Q_0e^{-k*6300}\]and if we do some cancellation of identical factors from both sides: \[\frac{1}{2}\cancel{Q_0} = \cancel{Q_0}e^{-k*6300}\]\[\frac{1}{2} = e^{-6300k}\]which you should be able to solve for \(k\).
Okay can you show me how to? Im confused

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Of course! What happens if we take the logarithm of an exponential?
\[\log_b b^x = x\]right? and \(\ln\) is just the special case of the logarithm to base \(e\), the base of the natural logarithm, \(e\approx 2.718281828...\)
That means if we take the natural logarithm of both sides: \[\ln(\frac{1}{2}) = \ln(e^{-6300k})\]\[\ln(\frac{1}{2}) = -6300k\]divide both sides by -6300 and you have your answer.
So when I came up with the answer I went to select it and it said it was actually -ln(0.5)/-6300k
the decay constant for Pu240 is \[k = -\frac{\ln{(\frac{1}{2}})}{6300} \approx 0.00011\]
You can check it by plugging it into the equation: \[Q(0) = Q_0*e^{-0.00011(0)} = Q_0*e^0 = Q_0\]\[Q(6300)=Q_0*e^{-0.00011*6300} = Q_0*e^{-0.693} = Q_0*0.5\]so after 6300 years (one half-life), we have half of the initial quantity.

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