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This problem involves continuous compounding, which is described by the formula:

\[
A = P e^{rt}
\]

Where:
- $ A $ is the amount of money after time $ t $ (in this case, $15,000),
- $ P $ is the initial investment (here, $10,000),
- $ r $ is the interest rate (8% or 0.08),
- $ t $ is the number of years,
- $ e $ is Euler's number (approximately 2.71828).

We need to solve for $ t $, the number of years it will take for the investment to grow from $10,000 to $15,000.

\[
15000 = 10000 e^{0.08t}
\]

First, divide both sides by 10,000:

\[
1.5 = e^{0.08t}
\]

Now, take the natural logarithm (ln) of both sides to eliminate the exponential:

\[
\ln(1.5) = 0.08t
\]

Finally, solve for $ t $:

\[
t = \frac{\ln(1.5)}{0.08}
\]

Let me calculate the value of $ t $.The investment will take approximately **5.07 years** to grow from $10,000 to $15,000 at an 8% continuously compounded interest rate.

Do you want further details or have any other questions?

Here are 5 related questions to deepen your understanding:
1. How does continuous compounding differ from annual compounding?
2. What would be the time if the interest rate was 6% instead of 8%?
3. How does the final amount change if the initial investment is $20,000?
4. What is the effect of doubling the initial investment on the growth time?
5. How would the calculation change if the final amount was $20,000?

**Tip**: Continuous compounding grows investments faster than periodic compounding because it accrues interest at every moment in time.

How many years will it take for an initial investment of $10,000 to grow to $15,000? Assume a rate of interest of 8% compounded continuously.

Learn how to calculate the time required for an investment of $10,000 to grow to $15,000 with an 8% continuously compounded interest rate. Using the formula A = P * e^(rt), this example covers key concepts like exponential growth and natural logarithms. Suitable for high school students in Grades 10-12.

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