Madeline invested \( \$ 2,800 \) in an account paying an interest rate of \( 4.4 \% \) compounded annually. 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 \( \$ 6,100 \) ?

To find out how long it takes for Madeline's investment to reach \( \$ 6,100 \), we can use the formula for compound interest: \[ A = P(1 + r)^t \] 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 number of years the money is invested or borrowed for. For this problem: - \( A = 6100 \) - \( P = 2800 \) - \( r = 0.044 \) Now, we need to solve for \( t \): \[ 6100 = 2800(1 + 0.044)^t \] Dividing both sides by 2800 gives: \[ \frac{6100}{2800} = (1.044)^t \] Calculating the left side: \[ 2.17857 \approx (1.044)^t \] Now, to solve for \( t \), we can take the logarithm of both sides: \[ \log(2.17857) = t \cdot \log(1.044) \] Now, solving for \( t \): \[ t = \frac{\log(2.17857)}{\log(1.044)} \] Using a calculator, we find: \[ t \approx \frac{0.3386}{0.0189} \approx 17.9 \] So, it would take approximately **17.9 years** for the value of the account to reach \( \$ 6,100 \).