Chapter 3 - Section 3.5 - Exponential Growth and Decay; Modeling Data - Exercise Set - Page 506: 26

$ 0.6134\%$ per hour = $-0.006134$

Model: $ A=A_{0}e^{kt}$ Unknown: $ k $, when $ A=0.5A_{0},\ t=113$ $0.5A_{0}=A_{0}e^{k\cdot 113}\qquad... /\div A_{0}$ $0.5=e^{k\cdot 113}\qquad.../\ln(...)$ $-0.693147=113k\qquad.../\div(113)$ $ k\approx-0.006134$ The decay rate is $ 0.6134\%$ per hour = $-0.006134$