Part 3: Compound Interest ( 10 points) A fast-food restaurant manager has saved \( \$ 15,000 \) for investment and is planning to invest the money in a retirement savings account offering \( 5.95 \% \) interest, compounded monthly. The amount of time it would take the manager to reach a balance of \( \$ 65,000 \) can be modeled by the equation log \( 4.33333=\log (1.00495833)^{12 t} \). Solve for the time in years, rounded to the nearest tenth of a year, showing all steps.

To solve for the time \( t \) in years, we start with the equation provided: \[ \log 4.33333 = \log (1.00495833)^{12t} \] Using the logarithmic power rule, we can move the exponent in front: \[ \log 4.33333 = 12t \cdot \log(1.00495833) \] Next, we isolate \( t \) by dividing both sides by \( 12 \cdot \log(1.00495833) \): \[ t = \frac{\log 4.33333}{12 \cdot \log(1.00495833)} \] Now, we will calculate \( \log 4.33333 \) and \( \log(1.00495833) \): 1. Using a calculator: - \( \log 4.33333 \approx 0.6386 \) - \( \log(1.00495833) \approx 0.00215 \) Substituting these values into the equation for \( t \): \[ t = \frac{0.6386}{12 \cdot 0.00215} \] Calculating the denominator: \[ 12 \cdot 0.00215 \approx 0.0258 \] Now we can calculate \( t \): \[ t \approx \frac{0.6386}{0.0258} \approx 24.752 \] Rounding to the nearest tenth: \[ t \approx 24.8 \] Thus, the manager will take approximately \( 24.8 \) years to reach a balance of \( \$65,000 \).