Mia invested \( \$ 52,000 \) in an account paying an interest rate of \( 3.1 \% \) compounded continuously. Assuming no deposits or withdrawals are made, how long would it take, to the nearest tenth of a year, for the value of the account to reach \( \$ 62,100 \) ?

To find out how long it would take for Mia's investment to reach \( \$62,100 \), we can use the formula for continuous compounding, which is: \[ A = Pe^{rt} \] Where: - \( A \) is the amount of money accumulated after time \( t \), - \( P \) is the principal amount (the initial investment), - \( r \) is the annual interest rate (as a decimal), - \( t \) is the time in years, - \( e \) is the base of the natural logarithm. Given: - \( P = 52,000 \) - \( A = 62,100 \) - \( r = 0.031 \) Plugging in the values, we have: \[ 62,100 = 52,000 e^{0.031t} \] Now, divide both sides by 52,000: \[ \frac{62,100}{52,000} = e^{0.031t} \] This simplifies to: \[ 1.19423 = e^{0.031t} \] Next, take the natural logarithm of both sides: \[ \ln(1.19423) = 0.031t \] Calculate \( \ln(1.19423) \): \[ \ln(1.19423) \approx 0.1781 \] Now, solve for \( t \): \[ t = \frac{0.1781}{0.031} \approx 5.743 \] To the nearest tenth of a year, it would take approximately \( 5.7 \) years for Mia's investment to reach \( \$62,100 \).