Eduardo invests 8,936 in a savings account with a fixed ammual interest rate of 3% compounded continuously . What will the account balance be after 6 years?

To solve this problem, we need to use the formula for continuous compounding interest:<br /><br />\[ A = P \cdot e^{rt} \]<br /><br />Where:<br />- \( A \) is the final amount (account balance)<br />- \( P \) is the principal amount (initial investment)<br />- \( r \) is the annual interest rate (as a decimal)<br />- \( t \) is the time in years<br />- \( e \) is the base of the natural logarithm (approximately 2.71828)<br /><br />Given:<br />- \( P = \$8,936 \)<br />- \( r = 3\% = 0.03 \)<br />- \( t = 6 \) years<br /><br />Now, plug these values into the formula:<br /><br />\[ A = 8936 \cdot e^{0.03 \cdot 6} \]<br /><br />First, calculate the exponent:<br /><br />\[ 0.03 \cdot 6 = 0.18 \]<br /><br />Next, calculate \( e^{0.18} \):<br /><br />\[ e^{0.18} \approx 1.197217 \]<br /><br />Now, multiply this by the principal amount:<br /><br />\[ A = 8936 \cdot 1.197217 \approx 10700.00 \]<br /><br />So, the account balance after 6 years will be approximately $\$10,700.00$.