Chapter 7 - Exponential Functions - 7.4 Exponential Growth and Decay - Exercises - Page 349: 14

$ k\approx 0.071$

Since the exponential decay is given by $$ P(t)=P_0 e^{-kt}, \quad k\gt 0$$ then we have $$ P(t)=10 e^{-kt}.$$ At $ t=17$, $ P(17)=3$, so: $$ P(17)=10 e^{-17k}=3\Longrightarrow k=-\frac{\ln(3/10)}{17}\approx 0.071.$$