Kaylee invested $830 in an account paying an interest rate of 6.6% compounded monthly. Assuming no deposits or withdrawals are made, how long would it take, *to the nearest year*, for the value of the account to reach $1,260?

To solve this problem, we will use the formula for compound interest:

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

where:

- $A$ is the amount of money in the account after time $t$,
- $P$ is the principal (initial investment),
- $r$ is the annual interest rate (as a decimal),
- $n$ is the number of times the interest is compounded per year,
- $t$ is the time in years.

Given in the problem:

- $P = 830$ (initial investment),
- $A = 1260$ (final amount),
- $r = 6.6\% = 0.066$ (annual interest rate),
- $n = 12$ (compounded monthly).

We want to find $t$, the time in years it takes for the account to reach $1,260.

First, substitute the given values into the compound interest formula:

\[
1260 = 830 \left(1 + \frac{0.066}{12}\right)^{12t}
\]

Simplify inside the parentheses:

\[
1260 = 830 \left(1 + 0.0055\right)^{12t}
\]

\[
1260 = 830 \cdot (1.0055)^{12t}
\]

Divide both sides by 830:

\[
\frac{1260}{830} = (1.0055)^{12t}
\]

Calculate the left side:

\[
1.5181 \approx (1.0055)^{12t}
\]

Now, take the natural logarithm (ln) of both sides to solve for $t$:

\[
\ln(1.5181) = \ln((1.0055)^{12t})
\]

Using the logarithmic property $\ln(a^b) = b \ln(a)$:

\[
\ln(1.5181) = 12t \cdot \ln(1.0055)
\]

Solve for $t$:

\[
t = \frac{\ln(1.5181)}{12 \cdot \ln(1.0055)}
\]

Calculate the values:

\[
t \approx \frac{\ln(1.5181)}{12 \times \ln(1.0055)} \approx \frac{0.4172}{0.0657} \approx 6.35
\]

Rounding to the nearest year:

\[
t \approx 6
\]

**Therefore, it would take approximately 6 years for Kaylee's investment to grow to $1,260.**

---

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. What if the interest was compounded quarterly instead of monthly? How would the time change?
2. How much would Kaylee's investment be worth after 10 years with the same interest rate?
3. What initial investment would be needed to reach $1,260 in 4 years at the same interest rate?
4. How does changing the interest rate to 7% affect the time needed to reach $1,260?
5. If Kaylee made a yearly deposit of $100, how long would it take to reach $1,260?

**Tip:** Remember, more frequent compounding generally increases the amount of interest earned over time, reducing the time needed to reach a certain investment goal.

Learn how to calculate compound interest with monthly compounding using a detailed example. Kaylee invested $830 at an interest rate of 6.6% compounded monthly, aiming to reach $1,260. Find out how long it takes, rounded to the nearest year.

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