## Calculus (3rd Edition)

$k\approx 0.071$
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: $$P(17)=10 e^{-17k}=3\Longrightarrow k=-\frac{\ln(3/10)}{17}\approx 0.071.$$