## Precalculus (6th Edition) Blitzer

$0.0428\%$ per year = $-0.000428$
Model: $A=A_{0}e^{kt}$ Unknown: $k$, when $A=0.5A_{0},\ t=1620$ $0.5A_{0}=A_{0}e^{k\cdot 1620}\qquad... /\div A_{0}$ $0.5=e^{k\cdot 1620}\qquad.../\ln(...)$ $-0.693147=1620k\qquad.../\div(1620)$ $k\approx-0.000428$ The decay rate is $0.0428\%$ per year = $-0.000428$