According to a census taken towards the end of the year 2009, the population of a rural town was found to be 64,000. The census uthority also found that the population of this particular town had a growth of 5% per annum. In how many years after 2009 did the population of this town reach 74,088 ?

Answer

To solve the problem of determining how many years it will take for the population of the town to grow from 64,000 to 74,088 with an annual growth rate of 5%, we can use the formula for compound interest:

\[ A = P \left(1 + \frac{r}{100}\right)^t \]

Where:
- \( A \) is the final amount (population after \( t \) years),
- \( P \) is the initial amount (initial population),
- \( r \) is the rate of growth (in percentage),
- \( t \) is the time in years.

### Step-by-Step Solution:

1. **Identify the values:**
- Initial population, \( P = 64,000 \)
- Final population, \( A = 74,088 \)
- Growth rate, \( r = 5\% \)

2. **Set up the equation using the formula:**
\[ 74,088 = 64,000 \left(1 + \frac{5}{100}\right)^t \]

3. **Simplify the growth factor:**
\[ 1 + \frac{5}{100} = 1 + 0.05 = 1.05 \]
So, the equation becomes:
\[ 74,088 = 64,000 (1.05)^t \]

4. **Divide both sides by 64,000:**
\[ \frac{74,088}{64,000} = (1.05)^t \]
Calculate the left side:
\[ \frac{74,088}{64,000} = 1.157 \]
Thus, we have:
\[ 1.157 = (1.05)^t \]

5. **Take the logarithm of both sides:**
\[ \log(1.157) = t \cdot \log(1.05) \]

6. **Solve for \( t \):**
\[ t = \frac{\log(1.157)}{\log(1.05)} \]
Using a calculator to find the logarithms:
- \( \log(1.157) \approx 0.069 \)
- \( \log(1.05) \approx 0.021 \)

Therefore:
\[ t \approx \frac{0.069}{0.021} \approx 3.29 \]

7. **Conclusion:**
Since \( t \) must be a whole number, we round \( 3.29 \) to 3. Thus, it will take approximately **3 years** for the population to reach 74,088.