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The problem asks to find how long it will take for an investment of $10,000 to grow to $20,000 when the interest is compounded continuously at a rate of 6% per year.

We can use the formula for continuous compound interest:

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

Where:
- $ A $ is the final amount ($20,000),
- $ P $ is the initial principal ($10,000),
- $ r $ is the annual interest rate (6% or 0.06),
- $ t $ is the time in years,
- $ e $ is the base of the natural logarithm.

### Step 1: Plug values into the formula.
\[
20000 = 10000 \cdot e^{0.06t}
\]

### Step 2: Simplify the equation.
\[
2 = e^{0.06t}
\]

### Step 3: Take the natural logarithm (ln) of both sides.
\[
\ln(2) = 0.06t
\]

### Step 4: Solve for $ t $.
\[
t = \frac{\ln(2)}{0.06}
\]

\[
t \approx \frac{0.6931}{0.06} \approx 11.55 \text{ years}
\]

### Conclusion:
The closest answer is **A) 11.5 years**.

Would you like more details, or do you have any questions?

### Related Questions:
1. What happens if the interest rate is reduced to 5%?
2. How does the continuous compounding formula differ from annual compounding?
3. What would the formula look like if the amount had tripled instead of doubled?
4. What if the initial amount was $5,000 instead of $10,000 with the same rate?
5. Can the natural logarithm be used to solve for any other variables?

### Tip:
For continuous compounding problems, always remember the base $ e $ is crucial for correct exponential growth calculations!

The current balance on an account with an initial principal of $10,000 is $20,000. The principal was invested at 6% annually, compounded continuously. Which of the following is closest to the amount of time the original $10,000 has been invested?

Learn how to calculate the time required for an investment of $10,000 to double to $20,000 with a 6% interest rate, using the continuous compounding formula. The problem involves concepts like exponential growth and natural logarithms. Suitable for students in Grades 10-12.

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