Find the time it takes for $6,000 to double when invested at an annual interest rate of 16%, compounded continuously.
 years  

To find the time it takes for an investment to double with continuous compounding, we can use the formula for continuously compounded interest:

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

Where:
- $ A $ is the final amount (in this case, $ 12,000 $ since the amount doubles),
- $ P $ is the initial principal (in this case, $ 6,000 $),
- $ r $ is the annual interest rate (in decimal form, 0.16 for 16%),
- $ t $ is the time in years,
- $ e $ is Euler's number (approximately 2.71828).

We want to find $ t $ when the amount doubles, so $ A = 2P $. Substituting this into the equation:

\[
2P = P e^{rt}
\]

Dividing both sides by $ P $:

\[
2 = e^{rt}
\]

Taking the natural logarithm of both sides:

\[
\ln(2) = rt
\]

Solving for $ t $:

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

Substitute $ r = 0.16 $:

\[
t = \frac{\ln(2)}{0.16}
\]

Now, calculating $ t $:

\[
t = \frac{0.6931}{0.16} \approx 4.33
\]

So, it will take approximately **4.33 years** for the investment to double.

Would you like further details on this, or have any questions?

Here are some related questions:
1. How would the time change if the interest rate were 12%?
2. What if the interest were compounded quarterly instead of continuously?
3. How does continuous compounding compare to simple interest for doubling an investment?
4. What is the formula for continuously compounded interest when finding the rate?
5. How can we calculate the future value for an investment compounded daily?

**Tip**: Continuous compounding leads to faster growth compared to other compounding methods due to its constant rate of application.

Find the time it takes for $6,000 to double when invested at an annual interest rate of 16%, compounded continuously.

Discover how long it takes for $6,000 to double at a 16% annual interest rate with continuous compounding. This problem involves exponential growth and logarithms, offering insights into continuously compounded interest. Suitable for students in Grades 10-12.

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