An engaged couple agrees to get married once their savings equal $57,250.00. They have $24,500.00 at the moment and will invest the money in an account that earns 8.00% per year. How many years will the engagement last?

1 answer

To determine how many years the engagement will last, we can use the formula for compound interest, which is:

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

Where:

  • \(A\) is the future value of the investment (the amount they want to have).
  • \(P\) is the principal amount (the initial amount of money, which is the current savings).
  • \(r\) is the annual interest rate (as a decimal).
  • \(t\) is the number of years the money is invested.

In this case:

  • \(A = 57,250\)
  • \(P = 24,500\)
  • \(r = 8% = 0.08\)

We want to solve for \(t\):

\[ 57,250 = 24,500(1 + 0.08)^t \]

First, we divide both sides by 24,500:

\[ \frac{57,250}{24,500} = (1.08)^t \]

Calculating the left side:

\[ \frac{57,250}{24,500} \approx 2.3333 \]

So we have:

\[ 2.3333 = (1.08)^t \]

To solve for \(t\), we take the logarithm of both sides:

\[ \log(2.3333) = \log((1.08)^t) \]

Using the power rule of logarithms:

\[ \log(2.3333) = t \cdot \log(1.08) \]

Now we can solve for \(t\):

\[ t = \frac{\log(2.3333)}{\log(1.08)} \]

Using a calculator, we find:

\[ \log(2.3333) \approx 0.3700 \] \[ \log(1.08) \approx 0.0334 \]

Now, substituting these values:

\[ t = \frac{0.3700}{0.0334} \approx 11.08 \]

Thus, the engagement will last approximately 11 years.