5100 dollars is placed in an account with an annual interest rate of \( 7 \% \). To the nearest year, how long will it take for the account value to reach 24000 dollars?

To solve this problem, we can use the formula for compound interest, which is given by: \[ A = P(1 + r)^t \] where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial amount of money). - \( r \) is the annual interest rate (decimal). - \( t \) is the time the money is invested or borrowed for, in years. In this scenario: - \( A = 24000 \) - \( P = 5100 \) - \( r = 0.07 \) Rearranging the formula to solve for \( t \): \[ 24000 = 5100(1 + 0.07)^t \] Dividing both sides by 5100: \[ \frac{24000}{5100} = (1.07)^t \] Calculating \( \frac{24000}{5100} \): \[ \approx 4.70588 \] Now we take the logarithm of both sides to solve for \( t \): \[ \log(4.70588) = t \cdot \log(1.07) \] Now, calculate \( \log(4.70588) \) and \( \log(1.07) \): \[ t = \frac{\log(4.70588)}{\log(1.07)} \] Calculating these logarithms (approximately): \[ \log(4.70588) \approx 0.6720 \quad \text{and} \quad \log(1.07) \approx 0.0291 \] Thus: \[ t \approx \frac{0.6720}{0.0291} \approx 23.06 \] Rounding to the nearest year, it would take approximately **23 years** for the account value to reach 24000 dollars.