To solve this problem, we’ll use the formula for continuous compounding:

$A = P e^{rt}$

where:

• $A$ is the amount after $t$ years,
• $P$ is the initial investment,
• $r$ is the annual interest rate,
• $t$ is the time in years.

Step 1: Time to Double

To find the time required for the investment to double, we set $A = 2P$:

$2P = P e^{rt}$

Dividing both sides by $P$:

$2 = e^{rt}$

Taking the natural logarithm of both sides:

$\ln(2) = rt$

Solving for $t$:

$t = \frac{\ln(2)}{r}$

Given $r = 8.8\% = 0.088$:

$t = \frac{\ln(2)}{0.088} \approx 7.88 \text{ years}$

So, the time to double is approximately 7.88 years.

Step 2: Amount After 10 Years

Now, to find the amount after 10 years, we substitute $t = 10$ into the formula:

$A = 1000 \times e^{0.088 \times 10}$

Calculating the exponent:

$A = 1000 \times e^{0.88} \approx 1000 \times 2.41 = 2410.14$

So, the amount after 10 years is approximately $2410.14.

Summary of Results:

• Time to Double: 7.88 years
• Amount After 10 Years: $2410.14

Would you like further explanation or details on these calculations?

Here are some related questions for deeper exploration:

1. What would be the time to double if the interest rate were 5% instead?
2. How would the amount change after 15 years at an 8.8% rate?
3. What is the effect of continuously compounded interest compared to annual compounding?
4. How is the formula for continuous compounding derived from the basic interest formula?
5. How does changing the initial investment affect the time to double?

Tip: Continuous compounding can yield significantly higher amounts over time compared to simple compounding methods, especially at higher interest rates.