ework 13: Applications of
Question 17, 4.5.32
HW Score: $44.12 \%, 7.5$ of 17 points
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garit
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A bird species in danger of extinction has a population that is decreasing exponentially $\left(A=A_{0} e^{k t}\right)$. Seven years ago the population was at 1800 and today only 1000 of the birds are alive. Once the population drops below 200 , the situation will be irreversible. How many years from now will this happen?

The population will drop below 200 birds approximately $\square$ years from now.
(Do not round until the final answer. Then round to the nearest whole number as needed.)
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Step 1 ：Given that the population of the bird species is decreasing exponentially, we can use the formula \(A = A0 * e^{kt}\), where \(A\) is the final amount, \(A0\) is the initial amount, \(k\) is the rate of decrease, and \(t\) is the time.

Step 2 ：From the problem, we know that seven years ago the population was 1800 and today it is 1000. We can use this information to find the rate of decrease, \(k\).

Step 3 ：First, we set up the equation with the given information: \(1000 = 1800 * e^{7k}\)

Step 4 ：To solve for \(k\), we first divide both sides by 1800: \(\frac{1000}{1800} = e^{7k}\)

Step 5 ：Then, we take the natural logarithm of both sides to get rid of the exponential: \(\ln(\frac{1000}{1800}) = 7k\)

Step 6 ：Finally, we solve for \(k\): \(k = \frac{\ln(\frac{1000}{1800})}{7} \approx -0.0645\)

Step 7 ：Now that we have the rate of decrease, we can use it to find out when the population will drop below 200. We set up the equation: \(200 = 1000 * e^{kt}\)

Step 8 ：And solve for \(t\): \(\frac{200}{1000} = e^{kt}\)

Step 9 ：Then, we take the natural logarithm of both sides to get rid of the exponential: \(\ln(\frac{200}{1000}) = kt\)

Step 10 ：Finally, we solve for \(t\): \(t = \frac{\ln(\frac{200}{1000})}{k} \approx 16.8\)

Step 11 ：So, the population will drop below 200 birds approximately 17 years from now (rounded to the nearest whole number).

Step 12 ：\(\boxed{17}\) years is the final answer.