A bird species in danger of extinction has a population that is decreasing exponentially. Five years ago, the population was 1400, and today only 1000 of the birds are alive. Once the population drops below 100, the situation will be irreversible. When will this happen?

Given:
[tex]\[
\begin{array}{l}
1000 = 1400 \times e^{5k} \\
R = -\ln (10 / 14) \\
100 = 1400 \times e^{kt}
\end{array}
\][/tex]

Exact answer:
[tex]\[
t = \frac{\ln (1 / 14)}{-0.065}
\][/tex]

Nearest year answer:
34.22



Answer :

To find out when the bird population will drop below 100, we can use the concept of exponential decay. Here is a detailed, step-by-step solution:

### Step-by-Step Solution:

1. Understanding Exponential Decay:
The population of the birds is decreasing exponentially. This can be expressed using the exponential decay formula:
[tex]\[ P(t) = P_0 \cdot e^{kt} \][/tex]
Where:
- [tex]\( P(t) \)[/tex] is the population at time [tex]\( t \)[/tex].
- [tex]\( P_0 \)[/tex] is the initial population.
- [tex]\( k \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time.

2. Define Given Values:
- Initial population, [tex]\( P_0 = 1400 \)[/tex]
- Population after 5 years, [tex]\( P(5) = 1000 \)[/tex]
- Time passed, [tex]\( t = 5 \)[/tex]

3. Find the Decay Constant [tex]\( k \)[/tex]:
We know that after 5 years, the population is 1000. So we can set up the equation:
[tex]\[ 1000 = 1400 \cdot e^{5k} \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ \frac{1000}{1400} = e^{5k} \][/tex]
[tex]\[ \frac{5}{7} = e^{5k} \][/tex]
Taking the natural logarithm (ln) on both sides:
[tex]\[ \ln\left(\frac{5}{7}\right) = 5k \][/tex]
[tex]\[ k = \frac{1}{5} \cdot \ln\left(\frac{5}{7}\right) \][/tex]
This results in:
[tex]\[ k \approx -0.0673 \][/tex]

4. Using [tex]\( k \)[/tex] to Find When Population Drops Below 100:
We want to find the time [tex]\( t \)[/tex] when the population [tex]\( P(t) \)[/tex] drops to 100.
[tex]\[ 100 = 1400 \cdot e^{kt} \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ \frac{100}{1400} = e^{kt} \][/tex]
[tex]\[ \frac{1}{14} = e^{kt} \][/tex]
Taking the natural logarithm (ln) on both sides:
[tex]\[ \ln\left(\frac{1}{14}\right) = kt \][/tex]
[tex]\[ t = \frac{\ln\left(\frac{1}{14}\right)}{k} \][/tex]

5. Calculate the Exact Time [tex]\( t \)[/tex]:
Plugging in the value of [tex]\( k \)[/tex]:
[tex]\[ t = \frac{\ln\left(\frac{1}{14}\right)}{-0.0673} \][/tex]
This results in:
[tex]\[ t \approx 39.217 \][/tex]

6. Answer:
The exact time when the population will drop below 100 is approximately 39.217 years. When rounded to the nearest year, it will happen in 39 years.

So, the population of birds will drop below 100 in about 39 years.

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