Find the time required for an investment of 3,000 dollars to grow to 9,000 dollars at an interest rate of 5% per year, compounded monthly. Give your answer accurate to 2 decimal places.

To solve for the time required for an investment to grow, we use the formula for compound interest:

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

Where:
- $A$ is the final amount (\$9,000),
- $P$ is the initial principal (\$3,000),
- $r$ is the annual interest rate (5%, or 0.05),
- $n$ is the number of times interest is compounded per year (monthly, so $n = 12$),
- $t$ is the time in years (which we are solving for).

Now, substitute the known values:

\[
9000 = 3000 \left( 1 + \frac{0.05}{12} \right)^{12t}
\]

First, simplify the equation:

\[
9000 = 3000 \left( 1 + \frac{0.05}{12} \right)^{12t}
\]

Divide both sides by 3000:

\[
3 = \left( 1 + \frac{0.05}{12} \right)^{12t}
\]

Now calculate the monthly interest rate:

\[
1 + \frac{0.05}{12} = 1.00416667
\]

So the equation becomes:

\[
3 = (1.00416667)^{12t}
\]

Take the natural logarithm (ln) of both sides to solve for $t$:

\[
\ln(3) = \ln\left( (1.00416667)^{12t} \right)
\]

Using the logarithmic property $\ln(a^b) = b\ln(a)$, this simplifies to:

\[
\ln(3) = 12t \cdot \ln(1.00416667)
\]

Solve for $t$:

\[
t = \frac{\ln(3)}{12 \cdot \ln(1.00416667)}
\]

Let's calculate this.The time required for the investment to grow from \$3,000 to \$9,000 at an interest rate of 5% per year, compounded monthly, is approximately **22.02 years**.

Would you like further details or have any questions?

Here are five related questions you might find interesting:
1. What would be the time required if the interest rate were 6% instead of 5%?
2. How much would the investment be worth after 10 years at the same interest rate?
3. What is the formula for continuously compounded interest, and how would it affect this problem?
4. How does compounding frequency (e.g., quarterly vs. monthly) affect the growth of an investment?
5. If you wanted the investment to grow to \$12,000, how much time would it take?

**Tip**: When calculating interest growth, small changes in compounding frequency can have a significant effect on the final amount, especially over long periods.

Learn how to calculate the time required for a $3,000 investment to grow to $9,000 at a 5% annual interest rate, compounded monthly. This problem involves compound interest formulas, logarithmic properties, and exponential growth, suitable for high school students in Grades 10-12.

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