You have \( \$ 300,000 \) saved for retirement. Your account earns \( 7 \% \) interest. How much will you be able to pull out each month, if you want to be able to take withdrawals for 25 years? \( \$ \square \)

Let the monthly withdrawal be \( P \) and use the annuity formula \[ P = \frac{r \cdot PV}{1 - (1 + r)^{-N}} \] where - \( PV = \$300,\!000 \) (the present value), - \( r = \frac{0.07}{12} \) (the monthly interest rate), and - \( N = 25 \times 12 = 300 \) (the total number of withdrawals). **Step 1. Find the monthly interest rate** \[ r = \frac{0.07}{12} \approx 0.0058333 \] **Step 2. Substitute the values into the formula** \[ P = \frac{\frac{0.07}{12} \times 300,\!000}{1 - \left(1 + \frac{0.07}{12}\right)^{-300}} \] This simplifies to \[ P = \frac{300,\!000 \times 0.0058333}{1 - \left(1.0058333\right)^{-300}} \] **Step 3. Compute the numerator** \[ 300,\!000 \times 0.0058333 \approx 1750 \] **Step 4. Compute the denominator** First, calculate \(\left(1.0058333\right)^{300}\). Taking logarithms: \[ \ln\left(1.0058333\right) \approx 0.0058166, \quad 300 \times 0.0058166 \approx 1.745 \] Thus, \[ \left(1.0058333\right)^{300} \approx e^{1.745} \approx 5.723 \] Then, \[ \left(1.0058333\right)^{-300} \approx \frac{1}{5.723} \approx 0.1748 \] Now, the denominator is \[ 1 - 0.1748 \approx 0.8252 \] **Step 5. Compute \( P \)** \[ P \approx \frac{1750}{0.8252} \approx 2120 \] Thus, the monthly withdrawal amount is approximately \[ \$2120 \]