To solve this problem, we can use the formula for continuously compounded interest:

$A = P e^{rt}$

Where:

• $A$ is the amount of money accumulated after interest (in this case, $2,400),
• $P$ is the principal amount (in this case, $1,900),
• $r$ is the annual interest rate (which we need to solve for),
• $t$ is the time the money is invested or borrowed for (in this case, 6 years),
• $e$ is the base of the natural logarithm (approximately 2.71828).

Step 1: Substitute the known values into the formula.

$2400 = 1900 e^{6r}$

Step 2: Solve for $e^{6r}$.

$\frac{2400}{1900} = e^{6r}$

$\frac{2400}{1900} \approx 1.2632$

$1.2632 = e^{6r}$

Step 3: Take the natural logarithm of both sides.

$\ln(1.2632) = \ln(e^{6r})$

$\ln(1.2632) \approx 0.2341$

$6r = 0.2341$

Step 4: Solve for $r$.

$r = \frac{0.2341}{6}$

$r \approx 0.0390$

Step 5: Convert to percentage.

$r \times 100 = 3.90\%$

So, the interest rate required is approximately 3.90%.

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Would you like further details on how this calculation works, or do you have any questions? Here are some related questions to explore:

1. How does the formula for compound interest differ when interest is compounded annually versus continuously?
2. What happens to the required interest rate if the time period changes to 5 years?
3. How do you calculate the interest earned if the principal is $1,500 and the interest rate is 4% compounded continuously for 10 years?
4. What is the impact of increasing the investment amount on the required interest rate to reach a specific target?
5. How does increasing the time period for investment affect the interest rate needed to reach a target amount?

Tip: Continuous compounding leads to slightly higher interest amounts compared to other compounding methods, as interest is calculated and added continuously rather than at set intervals.