Answer
$ -0.0149%$ per year$=-0.000149$
Work Step by Step
Exponential growth and decay models are given by $A=A_{0}e^{kt}$
in which $t$ represents time,
$A_{0}$ is the amount present at $t=0$, and
$A$ is the amount present at time $t$.
If $k>0$, the model describes growth and $k$ is the growth rate.
If $k<0$, the model describes decay and $k$ is the decay rate
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We have a decay model, and we know
the time $(t=4560)$ it takes for $A_{o}$ to become $0.5A_{0}.$
We solve for k:
$0.5A_{o}=A_{o}e^{t(4560)}\displaystyle \qquad .../\times\frac{1}{A_{o}}$
$0.5=e^{4560k}\qquad .../$ take ln( ) of both sides
$\ln 0.5=4560k$
$ k=\displaystyle \frac{\ln 0.5}{4560}\approx-0.000149=-0.0149%$
The decay rate is
$ -0.0149%$ per year$=-0.000149$
