Amanda places a bottle of water inside a cooler. As the water cools, its temperature \( C(t) \) in degrees Celsius is given by the following function, where \( t \) is the number of minutes since the bottle was placed in the cooler. \[ C(t)=8+14 e^{-0.03 t} \] Amanda wants to drink the water when it reaches a temperature of 16 degrees Celsius. How many minutes should she leave it in the cooler? Round your answer to the nearest tenth, and do not round any intermediate computations.

We start with the temperature function: \[ C(t) = 8 + 14 e^{-0.03t} \] Amanda wants the temperature to be \( 16^\circ \)C. So we set: \[ 16 = 8 + 14 e^{-0.03t} \] Subtract \( 8 \) from both sides: \[ 8 = 14 e^{-0.03t} \] Divide both sides by \( 14 \): \[ \frac{8}{14} = e^{-0.03t} \quad \Longrightarrow \quad \frac{4}{7} = e^{-0.03t} \] Take the natural logarithm of both sides: \[ \ln\left(\frac{4}{7}\right) = \ln\left(e^{-0.03t}\right) \] This simplifies using the logarithm property \( \ln\left(e^x\right) = x \): \[ \ln\left(\frac{4}{7}\right) = -0.03t \] Solve for \( t \): \[ t = \frac{-\ln\left(\frac{4}{7}\right)}{0.03} \] Now, compute \( \ln\left(\frac{4}{7}\right) \) approximately (keeping intermediate values unrounded): - \( \ln\left(\frac{4}{7}\right) \approx \ln(4) - \ln(7) \approx 1.3863 - 1.9459 \approx -0.5596 \) Thus, \[ t \approx \frac{-(-0.5596)}{0.03} = \frac{0.5596}{0.03} \approx 18.6533 \] Rounding to the nearest tenth gives: \[ t \approx 18.7 \text{ minutes} \]