Solve the following equation for \( t \). Round value to 1 decimal place if needed. \( 6600=2000\left(1+\frac{0.075}{12}\right)^{12 t} \) \( t=\square \)

To solve for \( t \), we first simplify the equation. Start with the given equation: \[ 6600 = 2000 \left(1+\frac{0.075}{12}\right)^{12t} \] Divide both sides by 2000: \[ \frac{6600}{2000} = \left(1+\frac{0.075}{12}\right)^{12t} \] Calculating the left side gives: \[ 3.3 = \left(1+\frac{0.075}{12}\right)^{12t} \] Now, calculate \( 1 + \frac{0.075}{12} \): \[ 1 + \frac{0.075}{12} = 1 + 0.00625 = 1.00625 \] We now have: \[ 3.3 = (1.00625)^{12t} \] Take the natural logarithm of both sides: \[ \ln(3.3) = \ln((1.00625)^{12t}) \] Using the power rule of logarithms: \[ \ln(3.3) = 12t \cdot \ln(1.00625) \] Now solve for \( t \): \[ t = \frac{\ln(3.3)}{12 \cdot \ln(1.00625)} \] Calculating each part: 1. \( \ln(3.3) \approx 1.193 \) 2. \( \ln(1.00625) \approx 0.00623 \) Substituting these values in: \[ t \approx \frac{1.193}{12 \cdot 0.00623} \approx \frac{1.193}{0.07476} \approx 15.94 \] Rounding to one decimal place: \( t \approx 15.9 \) So the final answer is: \( t = 15.9 \)