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

$A = P e^{rt}$

Where:

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

Step 1: Verify the "Time to Double" calculation

The formula for time to double is derived from:

$2P = P e^{rt}$

Simplify:

$2 = e^{rt}$

Take the natural logarithm ($\ln$) of both sides:

$\ln(2) = rt$

Solve for $t$:

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

Given $r = 8.8\% = 0.088$:

$t = \frac{\ln(2)}{0.088} \approx \frac{0.693}{0.088} \approx 7.88 \, \text{years}$

This matches the given "Time to Double" value.

Step 2: Calculate the "Amount After 10 Years"

Using the compound interest formula $A = P e^{rt}$, we substitute:

• $P = 1000$,
• $r = 0.088$,
• $t = 10$:

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

Simplify:

$A = 1000 e^{0.88}$

Using $e^{0.88} \approx 2.411$:

$A \approx 1000 \cdot 2.411 \approx 2411.09$

Thus, the "Amount After 10 Years" is $2411.09.

Final Table:

Would you like more details on any of the steps? Here are five questions to deepen your understanding:

1. What is the significance of $e$ in the continuously compounded interest formula?
2. How does the interest rate affect the time to double?
3. Can you derive the formula for time to double for continuously compounded interest from scratch?
4. How does continuous compounding differ from simple or periodic compounding?
5. What practical applications involve continuous compounding in finance or other fields?

Tip: Always double-check your interest rate conversions from percentages to decimals to avoid errors in financial calculations.