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

$k\approx 0.071$

#### Work Step by Step

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

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