What rate of interest compounded annually is required to double an investment in 6years?
 
Round your percentage to 2 decimal places.

Provide your answer below:

%  (Round your percentage to $2$ decimal places. 

To find the rate of interest that would double an investment in 6 years with annual compounding, we can use the formula for compound interest:

\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]

Where:
- $ A $ is the final amount (in this case, double the initial investment, so $ A = 2P $)
- $ P $ is the initial investment
- $ r $ is the annual interest rate (what we want to find)
- $ n $ is the number of times the interest is compounded per year (since it's compounded annually, $ n = 1 $)
- $ t $ is the time in years (6 years in this case)

We can rewrite the formula as:

\[
2P = P \left(1 + \frac{r}{1}\right)^6
\]

Simplify:

\[
2 = (1 + r)^6
\]

Now, solve for $ r $:

\[
(1 + r)^6 = 2
\]

Take the sixth root of both sides:

\[
1 + r = 2^{\frac{1}{6}}
\]

Now, subtract 1 from both sides to solve for $ r $:

\[
r = 2^{\frac{1}{6}} - 1
\]

Let me calculate this value.The annual interest rate required to double the investment in 6 years is approximately **12.25%**.

Would you like a detailed explanation or have any other questions?

Here are 5 related questions to explore further:
1. How does the interest rate change if the time to double is 8 years instead of 6?
2. What if the interest were compounded quarterly instead of annually?
3. How does the formula change for continuous compounding?
4. Can this method be applied to find the time needed to triple the investment?
5. How does inflation affect the real return on investment with a given interest rate?

**Tip**: When interest is compounded more frequently than annually, the effective interest rate becomes higher.

What rate of interest compounded annually is required to double an investment in 6 years? Round your percentage to 2 decimal places.

Find the annual interest rate required to double an investment in 6 years using the compound interest formula. This solution explores exponential growth and provides step-by-step calculations. Ideal for students in Grades 9-12.

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