Answer
$t \approx 39.2$ $years$
Work Step by Step
$$A = A_{0}e^{kt}$$ Let $A$ represent the amount of birds alive and $A_{0}$ the amount of birds alive 5 years ago ($t$). We can find the value of $k$ as follows: $$1,000 = 1,400e^{5k}$$ $$\frac{5}{7} = e^{5k}$$ $$5k = \ln{\frac{5}{7}}$$ $$k = \frac{\ln \frac{5}{7}}{5}$$ To find the value of $t$ when the population drops below 100, we can do as follows: $$100 = 1,400e^{\frac{\ln \frac{5}{7}}{5}t} = 1,400(e^{\ln \frac{5}{7}})^{\frac{t}{5}} = 1,400 (\frac{5}{7})^{\frac{t}{5}}$$ $$\frac{100}{1,400} = \frac{1}{14} = 14^{-1} = (\frac{5}{7})^{\frac{t}{5}}$$ $$\log_{\frac{5}{7}}14^{-1} = \frac{t}{5}$$ $$-5\log_{\frac{5}{7}}14 = t \approx 39.2$$
