You have 900-grams of an an unknown radioactive substance that has been determined to decay according to D ( t ) = 900 e − 0.002415 ⋅ t D ( t ) = 900 e - 0.002415 ⋅ t where t t is in years. How long before half of the initial amount has decayed?

It will take __ years for half of the initial amount to decay. (Round to 1 decimal place)

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01-11-2017
Mathematics

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You have 900-grams of an an unknown radioactive substance that has been determined to decay according to D ( t ) = 900 e − 0.002415 ⋅ t D ( t ) = 900 e - 0.002415 ⋅ t where t t is in years. How long before half of the initial amount has decayed?

It will take __ years for half of the initial amount to decay. (Round to 1 decimal place)

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jdoe0001 jdoe0001 · Answer 1 · 2017-11-01T23:07:40+01:00

now, the initial amount is 900 grams, so half of that will be 450 grams.

so, how long will it be, for D(t) to turn to 450 grams?

[tex]\bf D(t)=900e^{-0.002415t}\implies 450=900e^{-0.002415t} \\\\\\ \cfrac{450}{900}=e^{-0.002415t}\implies \cfrac{1}{2}=e^{-0.002415t}\\\\ -------------------------------\\\\ \textit{Logarithm Cancellation Rules}\\\\ log_{{ a}}{{ a}}^x\implies x\qquad \qquad {{ a}}^{log_{{ a}}x}=x\\\\ -------------------------------\\\\[/tex]

[tex]\bf log_e\left( \frac{1}{2} \right)=log_e\left( e^{-0.002415t} \right)\implies log_e\left( \frac{1}{2} \right)=-0.002415t \\\\\\ \cfrac{ln\left( \frac{1}{2}\right)}{-0.002415}=t\implies 287.0174661 \approx t[/tex]