Beth invests 6,329 in a savings account with a fixed annual interest rate of 3% compounded 3 times per year. How long will it take for the account balance to reach 7,131.68 3 years 4 years 6 years 5 years

To solve this problem, we can use the formula for compound interest:<br /><br />\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]<br /><br />where:<br />- \( A \) is the amount of money accumulated after \( t \) years, including interest.<br />- \( P \) is the principal amount (the initial amount of money).<br />- \( r \) is the annual interest rate (decimal).<br />- \( n \) is the number of times that interest is compounded per year.<br />- \( t \) is the time the money is invested for in years.<br /><br />Given:<br />- \( P = 6329 \)<br />- \( r = 0.03 \)<br />- \( n = 3 \)<br />- \( A = 7131.)<br /><br />We need to find \( t \).<br /><br />First, plug the given values into the formula:<br /><br />\[ 7131.68 = 6329 \left(1 + \frac{0.03}{3}\right)^{3t} \]<br /><br />Simplify inside the parentheses:<br /><br />\[ 7131.68 = 6329 \left(1 + 0.01\right)^{3t} \]<br />\[ 7131.68 = 6329 \left(1.01\right)^{3t} \]<br /><br />Next, divide both sides by 6329 to isolate the exponential term:<br /><br />\[ \frac{7131.68}{6329} = \left(1.01\right)^{3t} \]<br />\[ 1.125 = \left(1.01\right)^{3t} \]<br /><br />To solve for \( t \), take the natural logarithm (ln) of both sides:<br /><br />\[ \ln(1.125) = \ln\left((1.01)^{3t}\right) \]<br /><br />Using the property of logarithms that \( \ln(a^b) = b \ln(a) \):<br /><br />\[ \ln(1.125) = 3t \cdot \ln(1.01) \]<br /><br />Now, solve for \( t \):<br /><br />\[ t = \frac{\ln(1.125)}{3 \cdot \ln(1.01)} \]<br /><br />Calculate the values:<br /><br />\[ \ln(1.125) \approx 0.122 \]<br />\[ \.01) \approx 0.00995 \]<br /><br />So,<br /><br />\[ t = \frac{0.122}{3 \cdot 0.00995} \]<br />\[ t \approx \frac{0.122}{0.02985} \]<br />\[ t \approx 4.08 \]<br /><br />Therefore, it will take approximately 4 years for the account balance to reach $7,131.68.<br /><br />The correct answer is:<br />4 years