Chapter 3 - Section 3.5 - Exponential Growth and Decay; Modeling Data - Exercise Set - Page 506: 24

$ 0.0152\%$ per year = $-0.000152$

Model: $ A=A_{0}e^{kt}$ Unknown: $ k $, when $ A=0.5A_{0},\ t=4560$ $0.5A_{0}=A_{0}e^{k\cdot 4560}\qquad... /\div A_{0}$ $0.5=e^{k\cdot 4560}\qquad.../\ln(...)$ $-0.693147=4560k\qquad.../\div(4560)$ $ k\approx-0.000152$ The decay rate is $ 0.0152\%$ per year = $-0.000152$