https://file.aiquickdraw.com/mathbot/file/17282579819475guoyj7m.jpg

The formula provided for continuous compounding is:

\[
A = Pe^{rt}
\]

Where:
- $ A $ is the accumulated amount,
- $ P $ is the principal (initial amount invested),
- $ r $ is the annual interest rate (as a decimal),
- $ t $ is the time in years,
- $ e $ is the base of the natural logarithm.

From the image:
- $ P = 5500 $,
- $ r = 8\% = 0.08 $,
- The accumulated amount $ A $ is double the initial investment, so $ A = 2 \times 5500 = 11000 $.

We are tasked with solving for $ t $, the time in years required to double the investment under continuous compounding. Let's rearrange the formula:

\[
A = Pe^{rt}
\]

Substituting the known values:

\[
11000 = 5500e^{0.08t}
\]

Dividing both sides by 5500:

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

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

\[
\ln(2) = 0.08t
\]

Solve for $ t $:

\[
t = \frac{\ln(2)}{0.08}
\]

Let me calculate this.The time required to double the investment under continuous compounding at an 8% annual interest rate is approximately $ t \approx 8.7 $ years (rounded to one decimal place).

Do you need any more details or clarifications on this?

Here are five related questions you might consider:
1. How does continuous compounding differ from annual compounding?
2. What effect would a higher interest rate have on the time required to double the investment?
3. How would the time change if the interest rate were 5% instead of 8%?
4. What is the formula for compounded interest if it's compounded annually rather than continuously?
5. How is the natural logarithm used in the context of continuous compounding?

**Tip:** The rule of 72 is a quick way to estimate the time required to double an investment. Divide 72 by the annual interest rate to get an approximate doubling time.

How long will it take for an investment of $5500 to double at an annual interest rate of 8% under continuous compounding?

Learn how to calculate the time it takes for an investment of $5500 to double at an 8% interest rate under continuous compounding. This problem involves the use of exponential growth formulas and natural logarithms. Suitable for high school students in Grades 9-12.

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