## Precalculus (6th Edition) Blitzer

$t \approx 12.5$ days
The exponential decay process can be written as: $N(t) N_0 e^{-\lambda t}$ ...(1) But $\lambda =\dfrac{\ln 2}{t_{1/2}}= 0.096$ We need to simply equation (1) to obtain an expression for $t$ when $N(t)=0.3 N_0$ Plug in the given data: $0.3 N_0=N_0 e^{-0.096} t$ $\ln 0.3 =\ln (e^{-0.096t})$ or, $\ln 0.3 =-0.096t$ Thus, $t \approx 12.5$ days