Answer
17,121.7 years
Work Step by Step
Decay model: $A=A_{0}e^{kt} \qquad(k<0)$
($A_{0}$ is the initial quantity, A is the quantity after decaying for time t)
We first find k to define the decay equation.
Knowing the time $(t=$7340$)$ it takes for $A_{o}$ to become $0.5A_{0}$,
solve for k:
$0.5A_{0}=A_{0}e^{k(7340)}\displaystyle \qquad .../\times\frac{1}{A_{o}}$
$0.5=e^{7340k}\qquad .../$ take ln( ) of both sides
$\ln 0.5=7340k$
$k=\displaystyle \frac{\ln 0.5}{7340}\approx-0.000094$
$A=A_0 e^{-0.000094t}$
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Now we have a full formula into which we can substitute
$ A=0.20A_{0},\quad$and solve for t:
$0.20A_{0}=A_{0}e^{-0.000094t}\displaystyle \qquad .../\times\frac{1}{A_{o}}$
$0.20A_{0}=A_{0}e^{-0.000094t}\qquad .../$ take ln( ) of both sides
$\ln 0.2=-0.000094t$
$ t=\displaystyle \frac{\ln 0.2}{-0.000094}\approx$17121.6799$\quad$ years.
(Round to one decimal place)
It will take 17,121.7 years.
