QuestionA bird species in danger of extinction has a population that is decreasing exponentially ( $\mathrm{A}=\mathrm{A}_{0} \mathrm{c}^{\mathrm{kt}}$ ). Six 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. (Round the final answer to the nearest whole number as raeded. Round all intermediate values to 3 decimal places as needed.)

Studdy Solution

STEP 1

1. The population of the bird species is decreasing exponentially.
2. The exponential decay model is given by $A = A_0 c^{kt}$ .
3. Six years ago, the population was 1800.
4. The current population is 1000.
5. The population becomes irreversible once it drops below 200.
6. We need to find the number of years from now when the population will drop below 200.

STEP 2

1. Determine the initial population and the current population.
2. Use the exponential decay formula to find the decay constant $k$ .
3. Calculate the time $t$ when the population drops below 200.
4. Determine the number of years from now.

STEP 3

Determine the initial population and the current population.
- Initial population $A_0 = 1800$ (six years ago). - Current population $A = 1000$ .

STEP 4

Use the exponential decay formula to find the decay constant $k$ .
The formula is $A = A_0 c^{kt}$ .
Substitute the known values into the formula:
$1000 = 1800 \cdot c^{6k}$
Solve for $c^{6k}$ :
$c^{6k} = \frac{1000}{1800}$ $c^{6k} = \frac{5}{9}$
Take the natural logarithm of both sides to solve for $k$ :
$\ln(c^{6k}) = \ln\left(\frac{5}{9}\right)$ $6k \ln(c) = \ln\left(\frac{5}{9}\right)$
Assuming $c = e$ , we have:
$6k = \ln\left(\frac{5}{9}\right)$ $k = \frac{\ln\left(\frac{5}{9}\right)}{6}$
Calculate $k$ to three decimal places:
$k \approx \frac{\ln(0.555)}{6} \approx -0.109$

STEP 5

Calculate the time $t$ when the population drops below 200.
Use the formula $A = A_0 c^{kt}$ with $A = 200$ :
$200 = 1800 \cdot e^{kt}$
Solve for $t$ :
$e^{kt} = \frac{200}{1800}$ $e^{kt} = \frac{1}{9}$
Take the natural logarithm of both sides:
$kt = \ln\left(\frac{1}{9}\right)$
Substitute the value of $k$ :
$t = \frac{\ln\left(\frac{1}{9}\right)}{-0.109}$
Calculate $t$ to three decimal places:
$t \approx \frac{-2.197}{-0.109} \approx 20.165$

STEP 6

Determine the number of years from now.
Since $t$ is the total time from the initial population of 1800, we need to subtract the 6 years that have already passed:
$t_{\text{from now}} = 20.165 - 6 \approx 14.165$
Round to the nearest whole number:
The population will drop below 200 birds approximately $\boxed{14}$ years from now.

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