Questions: How long will it take 3,000 to grow to 17,000 if it is invested at 7% compounded monthly?

Solution Steps

To solve this problem, we need to use the formula for compound interest, which is given by:

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

where:

• \( A \) is the amount of money accumulated after n years, including interest.
• \( P \) is the principal amount (initial investment).
• \( r \) is the annual interest rate (decimal).
• \( n \) is the number of times that interest is compounded per year.
• \( t \) is the time the money is invested for in years.

We need to solve for \( t \) when \( A = 17000 \), \( P = 3000 \), \( r = 0.07 \), and \( n = 12 \).

Rearrange the formula to solve for \( t \):

\[ t = \frac{\log(\frac{A}{P})}{n \cdot \log(1 + \frac{r}{n})} \]

We are given the following values:

• Principal amount, \( P = 3000 \)
• Future amount, \( A = 17000 \)
• Annual interest rate, \( r = 0.07 \)
• Number of times interest is compounded per year, \( n = 12 \)

The formula for compound interest is:

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

We need to solve for \( t \), the time in years. Rearranging the formula gives:

\[ t = \frac{\log\left(\frac{A}{P}\right)}{n \cdot \log\left(1 + \frac{r}{n}\right)} \]

Substitute the given values into the rearranged formula:

\[ t = \frac{\log\left(\frac{17000}{3000}\right)}{12 \cdot \log\left(1 + \frac{0.07}{12}\right)} \]

Calculate the value of \( t \):

\[ t \approx 24.8522 \]

Round the calculated time to the nearest tenth of a year:

\[ t \approx 24.9 \]

Final Answer