A person invests 7500 dollars in a bank. The bank pays \( 5.5 \% \) interest compounded annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 14300 dollars? \[ A=P\left(1+\frac{r}{n}\right)^{n t} \]

To solve the problem using the formula for compound interest \( A = P\left(1 + \frac{r}{n}\right)^{nt} \), we can plug in the values. Here, \( A = 14300 \), \( P = 7500 \), \( r = 0.055 \), and \( n = 1 \) (since interest is compounded annually). Rearranging the formula to solve for \( t \): 1. Set up the equation: \[ 14300 = 7500\left(1 + 0.055\right)^{1t} \] 2. Simplifying gives: \[ 14300 = 7500(1.055)^{t} \] 3. Divide both sides by 7500: \[ \frac{14300}{7500} = (1.055)^{t} \] \[ 1.9067 = (1.055)^{t} \] 4. Taking the natural logarithm: \[ \ln(1.9067) = t \cdot \ln(1.055) \] 5. Solve for \( t \): \[ t = \frac{\ln(1.9067)}{\ln(1.055)} \approx \frac{0.644}{0.053} \approx 12.1 \] So, the person must leave the money in the bank for approximately 12.1 years. Investing has become a pivotal part of financial planning, and understanding how compound interest works can significantly impact long-term savings. You’re essentially “earning interest on interest,” which means the sooner you start saving, the more you benefit! This concept was first popularized in the 17th century by mathematicians like Jacob Bernoulli, who believed that compound interest could demonstrate the concept of exponential growth. When working with compound interest problems, one common mistake is forgetting to convert percentages into decimal form. Always remember to divide the percentage rate by 100 (e.g., \(5.5\% = 0.055\)). Also, ensuring that the compounding periods and ‘t’ (time) units match is crucial. A little oversight here can lead to miscalculations!