Questions: Suppose 500 is invested at 6% annual interest compounded twice a year. When will the investment be worth 1000?

Solution Steps

To solve this problem, we need to use the formula for compound interest. The formula is:

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

where:

• \( A \) is the amount of money accumulated after n years, including interest.
• \( P \) is the principal amount (the initial amount of money).
• \( r \) 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 \( A = 1000 \), \( P = 500 \), \( r = 0.06 \), and \( n = 2 \).

1. Use the compound interest formula to set up the equation.
2. Solve for \( t \) by isolating \( t \) on one side of the equation.
3. Use logarithms to solve for \( t \).

We start with the compound interest formula:

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

where:

• \( A = 1000 \) (the amount we want to reach),
• \( P = 500 \) (the initial investment),
• \( r = 0.06 \) (the annual interest rate),
• \( n = 2 \) (the number of times interest is compounded per year).

We need to isolate \( t \). Rearranging the formula gives us:

\[ 1000 = 500 \left(1 + \frac{0.06}{2}\right)^{2t} \]

This simplifies to:

\[ 2 = \left(1.03\right)^{2t} \]

Taking the logarithm of both sides, we have:

\[ \log(2) = 2t \cdot \log(1.03) \]

Now, we can solve for \( t \):

\[ t = \frac{\log(2)}{2 \cdot \log(1.03)} \]

Calculating this gives us:

\[ t \approx 11.7249 \]

Final Answer