## College Algebra (6th Edition)

The decay model: $A=A_{0}e^{kt}$ ($k < 0 )$ We find k from the half life information $0.5A_{0}=A_{0}e^{k\cdot 36}\qquad/\div A_{0}$ $0.5=e^{36k}\qquad$ ... apply ln( ) to both sides $\ln 0.5=36k$ $k=\displaystyle \frac{\ln 0.5}{36}\approx$-0.0192540883489$\approx-0.0193$ The decay model is : $A=A_{0}e^{-0.0193t}$ We want t (in hours) when $A=0.90A_{0}$ $0.9A_{0}=A_{0}e^{-0.0193t}\qquad/\div A_{0}$ $0.9=e^{-0.0193t}\qquad$ ... apply ln( ) to both sides $\ln 0.9=-0.0193t\qquad \div(-0.0193)$ $t=\displaystyle \frac{\ln 0.9}{-0.0193}\approx$5.459094075548$\approx 5.5$ hours