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Sagot :
Certainly! Let's solve the problem step-by-step.
### Problem:
We are given that the population of a village grows at a rate of 5% per annum. We need to find out how many years it will take for the population to decrease from 400,000 to 46,305.
### Step-by-Step Solution:
1. Identify Known Variables:
- Initial population ([tex]\(P_0\)[/tex]): 400,000
- Final population ([tex]\(P_t\)[/tex]): 46,305
- Annual growth rate (r): 5% or 0.05
2. Exponential Growth Formula:
The general formula for population growth (decay in this case) is:
[tex]\[ P_t = P_0 \times (1 + r)^t \][/tex]
where:
- [tex]\(P_t\)[/tex] is the final population
- [tex]\(P_0\)[/tex] is the initial population
- [tex]\(r\)[/tex] is the annual growth rate (expressed as a decimal)
- [tex]\(t\)[/tex] is the number of years
3. Re-arrange the Formula to Solve for Time ([tex]\(t\)[/tex]):
To find the number of years, we need to isolate [tex]\(t\)[/tex]:
[tex]\[ \left( \frac{P_t}{P_0} \right) = (1 + r)^t \][/tex]
Taking the natural logarithm (ln) of both sides:
[tex]\[ \ln\left( \frac{P_t}{P_0} \right) = \ln\left( (1 + r)^t \right) \][/tex]
Using the property of logarithms [tex]\(\ln(a^b) = b \ln(a)\)[/tex]:
[tex]\[ \ln\left( \frac{P_t}{P_0} \right) = t \cdot \ln(1 + r) \][/tex]
Solving for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{\ln\left( \frac{P_t}{P_0} \right)}{\ln(1 + r)} \][/tex]
4. Substitute the Known Values:
[tex]\[ t = \frac{\ln\left( \frac{46,305}{400,000} \right)}{\ln(1.05)} \][/tex]
5. Calculate the Fraction Inside the Logarithm:
[tex]\[ \frac{46,305}{400,000} = 0.1157625 \][/tex]
6. Compute the Natural Logarithms:
- [tex]\(\ln(0.1157625)\)[/tex]
- [tex]\(\ln(1.05)\)[/tex]
7. Perform the Division:
- The computed values for natural logarithms give us:
[tex]\[ t = \frac{\ln(0.1157625)}{\ln(1.05)} \approx \frac{-2.158}{0.04879} \][/tex]
8. Final Calculation:
[tex]\[ t \approx -44.19 \][/tex]
### Conclusion:
The negative value indicates that going from a population of 400,000 to 46,305 at a growth rate of 5% per annum would have taken approximately 44.19 years in the past. Therefore, the number of years required is approximately 44.19 years.
### Problem:
We are given that the population of a village grows at a rate of 5% per annum. We need to find out how many years it will take for the population to decrease from 400,000 to 46,305.
### Step-by-Step Solution:
1. Identify Known Variables:
- Initial population ([tex]\(P_0\)[/tex]): 400,000
- Final population ([tex]\(P_t\)[/tex]): 46,305
- Annual growth rate (r): 5% or 0.05
2. Exponential Growth Formula:
The general formula for population growth (decay in this case) is:
[tex]\[ P_t = P_0 \times (1 + r)^t \][/tex]
where:
- [tex]\(P_t\)[/tex] is the final population
- [tex]\(P_0\)[/tex] is the initial population
- [tex]\(r\)[/tex] is the annual growth rate (expressed as a decimal)
- [tex]\(t\)[/tex] is the number of years
3. Re-arrange the Formula to Solve for Time ([tex]\(t\)[/tex]):
To find the number of years, we need to isolate [tex]\(t\)[/tex]:
[tex]\[ \left( \frac{P_t}{P_0} \right) = (1 + r)^t \][/tex]
Taking the natural logarithm (ln) of both sides:
[tex]\[ \ln\left( \frac{P_t}{P_0} \right) = \ln\left( (1 + r)^t \right) \][/tex]
Using the property of logarithms [tex]\(\ln(a^b) = b \ln(a)\)[/tex]:
[tex]\[ \ln\left( \frac{P_t}{P_0} \right) = t \cdot \ln(1 + r) \][/tex]
Solving for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{\ln\left( \frac{P_t}{P_0} \right)}{\ln(1 + r)} \][/tex]
4. Substitute the Known Values:
[tex]\[ t = \frac{\ln\left( \frac{46,305}{400,000} \right)}{\ln(1.05)} \][/tex]
5. Calculate the Fraction Inside the Logarithm:
[tex]\[ \frac{46,305}{400,000} = 0.1157625 \][/tex]
6. Compute the Natural Logarithms:
- [tex]\(\ln(0.1157625)\)[/tex]
- [tex]\(\ln(1.05)\)[/tex]
7. Perform the Division:
- The computed values for natural logarithms give us:
[tex]\[ t = \frac{\ln(0.1157625)}{\ln(1.05)} \approx \frac{-2.158}{0.04879} \][/tex]
8. Final Calculation:
[tex]\[ t \approx -44.19 \][/tex]
### Conclusion:
The negative value indicates that going from a population of 400,000 to 46,305 at a growth rate of 5% per annum would have taken approximately 44.19 years in the past. Therefore, the number of years required is approximately 44.19 years.
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