The annual interest​ rate, compounded​ annually, at which ​$520 must be invested for it to grow to ​$1,956.31 in 14 years is

We can use the formula for compound interest to find the annual interest rate. The formula is:

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

Where:
- $A$ is the future amount (\$1,956.31),
- $P$ is the initial principal (\$520),
- $r$ is the annual interest rate (which we need to find),
- $t$ is the time in years (14 years),
- $n$ is the number of times the interest is compounded per year (since it is compounded annually, $n = 1$).

For annually compounded interest, the formula simplifies to:

\[
A = P(1 + r)^t
\]

Now we substitute the known values:

\[
1,956.31 = 520(1 + r)^{14}
\]

To solve for $r$, we first divide both sides by 520:

\[
\frac{1,956.31}{520} = (1 + r)^{14}
\]

\[
3.76213 = (1 + r)^{14}
\]

Next, take the 14th root of both sides:

\[
(1 + r) = \sqrt[14]{3.76213}
\]

Now, subtract 1 to find $r$:

\[
r = \sqrt[14]{3.76213} - 1
\]

Let me calculate this.The annual interest rate at which \$520 must be invested to grow to \$1,956.31 in 14 years is approximately **9.93%**.

Would you like more details or have any questions?

Here are 5 related questions to explore:
1. How would the investment change if it were compounded semi-annually instead?
2. What would the future value be if the investment were made at a 10% rate over 14 years?
3. How much time would it take for the investment to double at 9.93% annual interest?
4. What is the difference between simple interest and compound interest in this scenario?
5. How does the principal affect the final amount at a given interest rate and time?

**Tip:** The more frequently interest is compounded, the more the investment grows, even if the rate stays the same!

Discover how to calculate the annual interest rate needed for a $520 investment to grow to $1,956.31 over 14 years, using the compound interest formula. This step-by-step guide covers exponential growth and algebraic techniques, suitable for high school students in Grades 10-12.

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